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9.2: Solve Quadratic Equations Using the Square Root Property

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Learning Objectives

By the end of this section, you will be able to:

• Solve quadratic equations of the form a x 2 = k a x 2 = k using the Square Root Property
• Solve quadratic equations of the form a ( x – h ) 2 = k a ( x – h ) 2 = k using the Square Root Property

Be Prepared 9.1

Before you get started, take this readiness quiz.

Simplify: 128 . 128 . If you missed this problem, review Example 8.13.

Be Prepared 9.2

Simplify: 32 5 32 5 . If you missed this problem, review Example 8.50.

Be Prepared 9.3

Factor: 9 x 2 − 12 x + 4 9 x 2 − 12 x + 4 . If you missed this problem, review Example 6.23.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 . Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax 2 . We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form a x 2 = k a x 2 = k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x 2 = 9.

We can easily use factoring to find the solutions of similar equations, like x 2 = 16 and x 2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x = 4 , x = −4 x = 4 , x = −4 and x = 5 , x = −5 . x = 5 , x = −5 .

But what happens when we have an equation like x 2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13) 2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

If n 2 = m , then n is a square root of m . If n 2 = m , then n is a square root of m .

Since these equations are all of the form x 2 = k , the square root definition tells us the solutions are the two square roots of k . This leads to the Square Root Property .

Square Root Property

If x 2 = k , then

x = k or x = − k or x = ± k . x = k or x = − k or x = ± k .

Notice that the Square Root Property gives two solutions to an equation of the form x 2 = k , the principal square root of k k and its opposite. We could also write the solution as x = ± k . x = ± k . We read this as x equals positive or negative the square root of k .

Now we will solve the equation x 2 = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x 2 = 7.

We cannot simplify 7 7 , so we leave the answer as a radical.

Example 9.1

How to solve a quadratic equation of the form ax 2 = k using the square root property.

Solve: x 2 − 50 = 0 . x 2 − 50 = 0 .

Solve: x 2 − 48 = 0 . x 2 − 48 = 0 .

Solve: y 2 − 27 = 0 . y 2 − 27 = 0 .

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a quadratic equation using the square root property.

• Step 1. Isolate the quadratic term and make its coefficient one.
• Step 2. Use Square Root Property.
• Step 3. Simplify the radical.
• Step 4. Check the solutions.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Example 9.2

Solve: 3 z 2 = 108 . 3 z 2 = 108 .

Solve: 2 x 2 = 98 . 2 x 2 = 98 .

Solve: 5 m 2 = 80 . 5 m 2 = 80 .

The Square Root Property states ‘If x 2 = k x 2 = k ,’ What will happen if k < 0 ? k < 0 ? This will be the case in the next example.

Example 9.3

Solve: x 2 + 72 = 0 x 2 + 72 = 0 .

Solve: c 2 + 12 = 0 . c 2 + 12 = 0 .

Solve: q 2 + 24 = 0 . q 2 + 24 = 0 .

Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Example 9.4

Solve: 2 3 u 2 + 5 = 17 . 2 3 u 2 + 5 = 17 .

Solve: 1 2 x 2 + 4 = 24 . 1 2 x 2 + 4 = 24 .

Solve: 3 4 y 2 − 3 = 18 . 3 4 y 2 − 3 = 18 .

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .

Example 9.5

Solve: 2 x 2 − 8 = 41 . 2 x 2 − 8 = 41 .

Solve: 5 r 2 − 2 = 34 . 5 r 2 − 2 = 34 .

Try It 9.10

Solve: 3 t 2 + 6 = 70 . 3 t 2 + 6 = 70 .

Solve Quadratic Equations of the Form a ( x − h ) 2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a ( x − h ) 2 = k as well. Notice that the quadratic term, x , in the original form ax 2 = k is replaced with ( x − h ).

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a , then the Square Root Property can be used on ( x − h ) 2 .

Example 9.6

Solve: 4 ( y − 7 ) 2 = 48 . 4 ( y − 7 ) 2 = 48 .

Try It 9.11

Solve: 3 ( a − 3 ) 2 = 54 . 3 ( a − 3 ) 2 = 54 .

Try It 9.12

Solve: 2 ( b + 2 ) 2 = 80 . 2 ( b + 2 ) 2 = 80 .

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example 9.7

Solve: ( x − 1 3 ) 2 = 5 9 . ( x − 1 3 ) 2 = 5 9 .

Try It 9.13

Solve: ( x − 1 2 ) 2 = 5 4 . ( x − 1 2 ) 2 = 5 4 .

Try It 9.14

Solve: ( y + 3 4 ) 2 = 7 16 . ( y + 3 4 ) 2 = 7 16 .

We will start the solution to the next example by isolating the binomial term.

Example 9.8

Solve: 2 ( x − 2 ) 2 + 3 = 57 . 2 ( x − 2 ) 2 + 3 = 57 .

Try It 9.15

Solve: 5 ( a − 5 ) 2 + 4 = 104 . 5 ( a − 5 ) 2 + 4 = 104 .

Try It 9.16

Solve: 3 ( b + 3 ) 2 − 8 = 88 . 3 ( b + 3 ) 2 − 8 = 88 .

Sometimes the solutions are complex numbers.

Example 9.9

Solve: ( 2 x − 3 ) 2 = −12 . ( 2 x − 3 ) 2 = −12 .

Try It 9.17

Solve: ( 3 r + 4 ) 2 = −8 . ( 3 r + 4 ) 2 = −8 .

Try It 9.18

Solve: ( 2 t − 8 ) 2 = −10 . ( 2 t − 8 ) 2 = −10 .

The left sides of the equations in the next two examples do not seem to be of the form a ( x − h ) 2 . But they are perfect square trinomials, so we will factor to put them in the form we need.

Example 9.10

Solve: 4 n 2 + 4 n + 1 = 16 . 4 n 2 + 4 n + 1 = 16 .

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

Try It 9.19

Solve: 9 m 2 − 12 m + 4 = 25 . 9 m 2 − 12 m + 4 = 25 .

Try It 9.20

Solve: 16 n 2 + 40 n + 25 = 4 . 16 n 2 + 40 n + 25 = 4 .

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

• Solving Quadratic Equations: The Square Root Property
• Using the Square Root Property to Solve Quadratic Equations

Section 9.1 Exercises

Practice makes perfect.

Solve Quadratic Equations of the Form ax 2 = k Using the Square Root Property

In the following exercises, solve each equation.

a 2 = 49 a 2 = 49

b 2 = 144 b 2 = 144

r 2 − 24 = 0 r 2 − 24 = 0

t 2 − 75 = 0 t 2 − 75 = 0

u 2 − 300 = 0 u 2 − 300 = 0

v 2 − 80 = 0 v 2 − 80 = 0

4 m 2 = 36 4 m 2 = 36

3 n 2 = 48 3 n 2 = 48

4 3 x 2 = 48 4 3 x 2 = 48

5 3 y 2 = 60 5 3 y 2 = 60

x 2 + 25 = 0 x 2 + 25 = 0

y 2 + 64 = 0 y 2 + 64 = 0

x 2 + 63 = 0 x 2 + 63 = 0

y 2 + 45 = 0 y 2 + 45 = 0

4 3 x 2 + 2 = 110 4 3 x 2 + 2 = 110

2 3 y 2 − 8 = −2 2 3 y 2 − 8 = −2

2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11

3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41

7 p 2 + 10 = 26 7 p 2 + 10 = 26

2 q 2 + 5 = 30 2 q 2 + 5 = 30

5 y 2 − 7 = 25 5 y 2 − 7 = 25

3 x 2 − 8 = 46 3 x 2 − 8 = 46

( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64

( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121

( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20

( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32

( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4

( x + 1 5 ) 2 = 7 25 ( x + 1 5 ) 2 = 7 25

( y + 2 3 ) 2 = 8 81 ( y + 2 3 ) 2 = 8 81

( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25

( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55

( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39

4 ( x + 3 ) 2 − 5 = 27 4 ( x + 3 ) 2 − 5 = 27

5 ( x + 3 ) 2 − 7 = 68 5 ( x + 3 ) 2 − 7 = 68

( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27

( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24

( 4 x − 3 ) 2 + 11 = −17 ( 4 x − 3 ) 2 + 11 = −17

( 2 y + 1 ) 2 − 5 = −23 ( 2 y + 1 ) 2 − 5 = −23

m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8

n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27

x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12

y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32

25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36

9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9

36 x 2 − 24 x + 4 = 81 36 x 2 − 24 x + 4 = 81

64 x 2 + 144 x + 81 = 25 64 x 2 + 144 x + 81 = 25

Mixed Practice

In the following exercises, solve using the Square Root Property.

2 r 2 = 32 2 r 2 = 32

4 t 2 = 16 4 t 2 = 16

( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28

( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8

9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1

4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49

a 2 − 18 = 0 a 2 − 18 = 0

b 2 − 108 = 0 b 2 − 108 = 0

( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9

( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4

m 2 + 12 = 0 m 2 + 12 = 0

n 2 + 48 = 0 . n 2 + 48 = 0 .

u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72

v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50

( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15

( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64

( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4

( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64

6 c 2 + 4 = 29 6 c 2 + 4 = 29

2 d 2 − 4 = 77 2 d 2 − 4 = 77

( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3

( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9

Writing Exercises

In your own words, explain the Square Root Property.

In your own words, explain how to use the Square Root Property to solve the quadratic equation ( x + 2 ) 2 = 16 ( x + 2 ) 2 = 16 .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

Square Root Property

What is Square Root Property？

The square root property is a pivotal concept in algebra for solving quadratic equations. The square root property simplifies the process of finding the roots of a quadratic equation. It is particularly useful because it gives a direct way to solve for $ x $ when the equation is already in the form of $ x^2 = a $, without the need for more complex methods.

The properties of square roots include:

• Product Property : $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, where $a \geq 0$ and $b \geq 0$.
• Quotient Property : $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, where $a \geq 0$ and $b > 0$.
• Square Root of a Square : $\sqrt{x^2} = |x|$, reflecting that the square root returns the non-negative root.
• Rationalizing the Denominator : Involves multiplying top and bottom by a term that eliminates the square root in the denominator.
• Even Roots of Negative Numbers : Not defined in the real number system. $√(-a)$ is imaginary if $a > 0$.

Using the Square Root Property

Detailed Steps to Solve Quadratic Equations Using the Square Root Property:

• Converting the equation: Convert the equation into the form that conforms to the square root Property.
• Identify the Numbers : Determine the values of a, b, and x.
• Apply the Square Root : Apply the square root to both sides of the equation, keeping in mind to include the positive and negative roots.
• Simplify the Equation : After applying the square root property, ensure the other side of the equation is a simplified constant. If there are other terms, add or subtract them.
• Solve for $ x $ : After simplifying the equation, you will get two possible solutions for $ x $: $ x = +\sqrt{a} $ and $ x = -\sqrt{a} $.
• Check Your Solutions : It's good practice to substitute your solutions back into the original equation to ensure they work.

Solved Examples

Example #1: using the product property of square roots.

Multiply $\sqrt{12}$ and $\sqrt{3}$.

Step 1: Identify the Numbers

Step 2: Apply the Product Property

• $\sqrt{12} \cdot \sqrt{3} = \sqrt{12 \cdot 3}$

Step 3: Calculate the Product

• $12 \cdot 3 = 36$

Step 4: Simplify the Square Root

• $\sqrt{36} = 6$
• $\sqrt{12} \cdot \sqrt{3} = 6$

Example #2: Using the Quotient Property

Simplify $\sqrt{\frac{50}{2}}$.

Step 2: Apply the Quotient Property

• $\sqrt{\frac{50}{2}} = \frac{\sqrt{50}}{\sqrt{2}}$

Step 3: Simplify Each Square Root (if possible)

• $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ (since $\sqrt{25} = 5$)
• $\sqrt{2}$ remains as it is because it cannot be simplified further.

Step 4: Form the Simplified Expression

• $\frac{5\sqrt{2}}{\sqrt{2}}$

Step 5: Simplify the Expression

• When you divide $5\sqrt{2}$ by $\sqrt{2}$, the $\sqrt{2}$ in the numerator and denominator cancel out, leaving $5$.
• $\sqrt{\frac{50}{2}} = 5$

Example #3: Using Square Root of a Square

Evaluate $\sqrt{(-5)^2}$.

Step 1: Square the Number

• $(-5)^2 = 25$.

Step 2: Apply the Square Root

• $\sqrt{25}$.

Step 3: Evaluate the Square Root

• $\sqrt{25} = 5$.

Step 4: Apply the Property

• Since $\sqrt{(-5)^2}$ results in $5$, but knowing $-5$ was the original number, we acknowledge $\sqrt{(-5)^2} = |{-5}|$.

Conclusion:

• $\sqrt{(-5)^2} = |{-5}| = 5$, which matches the result of step 3, confirming the property $\sqrt{x^2} = |x|$.

Example #4: Using Rationalizing the Denominator

Let's rationalize the denominator for the expression $\frac{3}{\sqrt{2}}$.

Step 1: Identify the Expression

• Fraction with a square root in the denominator: $\frac{3}{\sqrt{2}}$.

Step 2: Multiply by a Form of 1

• To rationalize, multiply both the numerator and denominator by the square root in the denominator: $\sqrt{2}$.
• Form of 1: $\frac{\sqrt{2}}{\sqrt{2}}$.

Step 3: Apply the Multiplication

• Multiply the numerator: $3 \times \sqrt{2} = 3\sqrt{2}$.
• Multiply the denominator: $\sqrt{2} \times \sqrt{2} = 2$.

Step 4: Write the Simplified Expression

• The resulting expression: $\frac{3\sqrt{2}}{2}$.
• The expression $\frac{3}{\sqrt{2}}$ is rationalized as $\frac{3\sqrt{2}}{2}$.

Example #5: Using Even Roots of Negative Numbers

Find the square root of $-16$.

Step 1: Express $-16$ as $16 \times -1$

• $-16 = 16 \times -1$.
• Since the square root of $-1$ is $i$, and the square root of $16$ is $4$, you take the square root of each factor separately:
• $\sqrt{-16} = \sqrt{16} \times \sqrt{-1}$.

Step 3: Simplify Using the Definition of $i$

• $\sqrt{16} = 4$ and $\sqrt{-1} = i$.

Step 4: Combine the Results

• $4 \times i = 4i$.
• The square root of $-16$ is $4i$, illustrating how to handle even roots of negative numbers by using imaginary numbers.

FAQ About Square Root Property

Q: what is difference between square and square root.

A: The difference between square and square root is:

Square : Multiplying a number by itself. Expressed as $n^2$. For $n=3$, square is $3^2 = 9$.

Square Root : Finding a number that, when multiplied by itself, gives the original number. Expressed as $\sqrt{n}$. For $n=9$, square root is $\sqrt{9} = 3$.

Q: What is the vocabulary of square root?

A: The vocabulary of square root includes several key terms:

Square Root : The square root of a number $x$ is a value that, when multiplied by itself, equals $x$. For example, the square root of 9 ($\sqrt{9}$) is 3, because $3 \times 3 = 9$.

Radicand : The number under the square root symbol. In $\sqrt{9}$, 9 is the radicand.

Radical Sign : The symbol $\sqrt{}$ used to denote the square root. For example, in $\sqrt{16}$, the symbol $\sqrt{}$, is the radical sign.

Principal Square Root : The non-negative square root of a number. Every positive number has two square roots: a positive and a negative. The principal square root refers to the positive one. For example, the principal square root of 16 is 4, not -4.

Perfect Square : A number that is the square of an integer. For instance, 16 is a perfect square because it’s $4^2$.

Rationalize : The process of adjusting an expression to eliminate square roots from the denominator. For example, to rationalize $\frac{1}{\sqrt{2}}$, you multiply both numerator and denominator by $\sqrt{2}$ to get $\frac{\sqrt{2}}{2}$.

Imaginary Unit ($i$) : Represented by $i$, it is defined as $\sqrt{-1}$. It’s used to express square roots of negative numbers, given that no real number squared equals a negative number. For instance, $\sqrt{-4} = 2i$.

Q: What number is both square and cube?

A: A number that is both a perfect square and a perfect cube is called a perfect sixth power. For example, $64$ is both a perfect square ($8^2$) and a perfect cube ($4^3$).

Q: How do you know if a number is irrational?

A: A number is irrational if it meets certain criteria, indicating it cannot be exactly written as a ratio of two integers $a/b$, where $a$ and $b$ are integers and $b \neq 0$. Key characteristics of irrational numbers include:

Non-Terminating, Non-Repeating Decimal Expansion : Irrational numbers have infinite decimal digits that do not terminate or form a repeating pattern.

Cannot Be Expressed as a Fraction : Unlike rational numbers, which can be written as fractions, irrational numbers cannot be precisely represented in fractional form.

Roots of Non-Perfect Squares (and other degrees) : The square roots (and roots of other degrees) of non-perfect squares are often irrational. For example, $\sqrt{2}$ cannot be represented as a finite or repeating decimal, nor as a fraction of two integers, making it irrational.

Example : Proving $\sqrt{2}$ is Irrational

To understand why $\sqrt{2}$ is irrational, consider if it were rational, it could be written as $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are coprime integers (their greatest common divisor is 1), and $b \neq 0$.

• Squaring both sides yields $2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$.
• This equation implies $a^2$ is even (since it's 2 times some number).
• If $a^2$ is even, then $a$ itself must be even (since the square of an odd number is odd).
• Let $a = 2k$ (where $k$ is an integer), then substituting back: $4k^2 = 2b^2$ or $2k^2 = b^2$, showing $b^2$, and thus $b$, must also be even.
• However, this contradicts our assumption that $a$ and $b$ are coprime, as both being even means they share at least a factor of $2$.

Therefore, no such integers $a$ and $b$ exist that can satisfy the condition $\sqrt{2} = \frac{a}{b}$, making $\sqrt{2}$ irrational. This method illustrates a fundamental way of identifying an irrational number through contradiction.

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Course: Algebra 1   >   Unit 14

Solving quadratics by taking square roots.

• Quadratics by taking square roots (intro)
• Solving quadratics by taking square roots examples
• Quadratics by taking square roots
• Quadratics by taking square roots: strategy
• Solving quadratics by taking square roots: with steps
• Quadratics by taking square roots: with steps
• Solving simple quadratics review

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Video transcript

9.6 Solve Equations with Square Roots

Learning objectives.

By the end of this section, you will be able to:

• Solve radical equations
• Use square roots in applications

Be Prepared 9.15

Before you get started, take this readiness quiz.

Simplify: ⓐ 9 9 ⓑ 9 2 9 2 . If you missed this problem, review Example 9.1 and Example 1.19 .

Be Prepared 9.16

Solve: 5 ( x + 1 ) − 4 = 3 ( 2 x − 7 ) 5 ( x + 1 ) − 4 = 3 ( 2 x − 7 ) . If you missed this problem, review Example 2.42 .

Be Prepared 9.17

Solve: n 2 − 6 n + 8 = 0 n 2 − 6 n + 8 = 0 . If you missed this problem, review Example 7.73 .

Solve Radical Equations

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

Radical Equation

An equation in which the variable is in the radicand of a square root is called a radical equation .

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write a a we mean the principal square root. So a ≥ 0 a ≥ 0 always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make a a negative. This algebraic solution would not be a solution to the original radical equation ; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.

Example 9.74

For the equation x + 2 = x x + 2 = x :

ⓐ Is x = 2 x = 2 a solution? ⓑ Is x = −1 x = −1 a solution?

ⓐ Is x = 2 x = 2 a solution?

ⓑ Is x = −1 x = −1 a solution?

Try It 9.147

For the equation x + 6 = x x + 6 = x :

ⓐ Is x = −2 x = −2 a solution? ⓑ Is x = 3 x = 3 a solution?

Try It 9.148

For the equation − x + 2 = x − x + 2 = x :

ⓐ Is x = −2 x = −2 a solution? ⓑ Is x = 1 x = 1 a solution?

Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring.

Example 9.75

How to solve radical equations.

Solve: 2 x − 1 = 7 2 x − 1 = 7 .

Try It 9.149

Solve: 3 x − 5 = 5 3 x − 5 = 5 .

Try It 9.150

Solve: 4 x + 8 = 6 4 x + 8 = 6 .

Solve a radical equation.

• Step 1. Isolate the radical on one side of the equation.
• Step 2. Square both sides of the equation.
• Step 3. Solve the new equation.
• Step 4. Check the answer.

Example 9.76

Solve: 5 n − 4 − 9 = 0 5 n − 4 − 9 = 0 .

Try It 9.151

Solve: 3 m + 2 − 5 = 0 3 m + 2 − 5 = 0 .

Try It 9.152

Solve: 10 z + 1 − 2 = 0 10 z + 1 − 2 = 0 .

Example 9.77

Solve: 3 y + 5 + 2 = 5 3 y + 5 + 2 = 5 .

Try It 9.153

Solve: 3 p + 3 + 3 = 5 3 p + 3 + 3 = 5 .

Try It 9.154

Solve: 5 q + 1 + 4 = 6 5 q + 1 + 4 = 6 .

When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.

Example 9.78

Solve: 9 k − 2 + 1 = 0 9 k − 2 + 1 = 0 .

Try It 9.155

Solve: 2 r − 3 + 5 = 0 2 r − 3 + 5 = 0 .

Try It 9.156

Solve: 7 s − 3 + 2 = 0 7 s − 3 + 2 = 0 .

If one side of the equation is a binomial, we use the binomial squares formula when we square it.

Binomial Squares

Don’t forget the middle term!

Example 9.79

Solve: p − 1 + 1 = p p − 1 + 1 = p .

Try It 9.157

Solve: x − 2 + 2 = x x − 2 + 2 = x .

Try It 9.158

Solve: y − 5 + 5 = y y − 5 + 5 = y .

Example 9.80

Solve: r + 4 − r + 2 = 0 r + 4 − r + 2 = 0 .

Try It 9.159

Solve: m + 9 − m + 3 = 0 m + 9 − m + 3 = 0 .

Try It 9.160

Solve: n + 1 − n + 1 = 0 n + 1 − n + 1 = 0 .

When there is a coefficient in front of the radical, we must square it, too.

Example 9.81

Solve: 3 3 x − 5 − 8 = 4 3 3 x − 5 − 8 = 4 .

Try It 9.161

Solve: 2 4 a + 2 − 16 = 16 2 4 a + 2 − 16 = 16 .

Try It 9.162

Solve: 3 6 b + 3 − 25 = 50 3 6 b + 3 − 25 = 50 .

Example 9.82

Solve: 4 z − 3 = 3 z + 2 4 z − 3 = 3 z + 2 .

Try It 9.163

Solve: 2 x − 5 = 5 x + 3 2 x − 5 = 5 x + 3 .

Try It 9.164

Solve: 7 y + 1 = 2 y − 5 7 y + 1 = 2 y − 5 .

Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.

Example 9.83

Solve: m + 1 = m + 9 m + 1 = m + 9 .

Try It 9.165

Solve: x + 3 = x + 5 x + 3 = x + 5 .

Try It 9.166

Solve: m + 5 = m + 16 m + 5 = m + 16 .

Example 9.84

Solve: q − 2 + 3 = 4 q + 1 q − 2 + 3 = 4 q + 1 .

Try It 9.167

Solve: y − 3 + 2 = 4 y + 2 y − 3 + 2 = 4 y + 2 .

Try It 9.168

Solve: n − 4 + 5 = 3 n + 3 n − 4 + 5 = 3 n + 3 .

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.

We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.

Solve applications with formulas.

• Step 1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for by choosing a variable to represent it.
• Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

We used the formula A = L · W A = L · W to find the area of a rectangle with length L and width W . A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is s 2 s 2 .

The formula A = s 2 A = s 2 gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s .

A = s 2 Take the square root of both sides. A = s 2 Simplify. A = s A = s 2 Take the square root of both sides. A = s 2 Simplify. A = s

We can use the formula s = A s = A to find the length of a side of a square for a given area.

Area of a Square

We will show an example of this in the next example.

Example 9.85

Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula s = A s = A to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.

Try It 9.169

Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. Use the formula s = A s = A to find the length of each side of her lawn. Round your answer to the nearest tenth of a foot.

Try It 9.170

Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. Use the formula s = A s = A to find the length of each side of his mosaic. Round your answer to the nearest tenth of a centimeter.

Another application of square roots has to do with gravity.

Falling Objects

On Earth, if an object is dropped from a height of h h feet, the time in seconds it will take to reach the ground is found by using the formula,

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h = 64 h = 64 into the formula.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Example 9.86

Christy dropped her sunglasses from a bridge 400 feet above a river. Use the formula t = h 4 t = h 4 to find how many seconds it took for the sunglasses to reach the river.

Try It 9.171

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the package to reach the ground.

Try It 9.172

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula t = h 4 t = h 4 to find how many seconds it took for the squeegee to reach the sidewalk.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

Skid Marks and Speed of a Car

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula,

Example 9.87

After a car accident, the skid marks for one car measured 190 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 9.173

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 9.174

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Section 9.6 Exercises

Practice makes perfect.

In the following exercises, check whether the given values are solutions.

For the equation x + 12 = x x + 12 = x : ⓐ Is x = 4 x = 4 a solution? ⓑ Is x = −3 x = −3 a solution?

For the equation − y + 20 = y − y + 20 = y : ⓐ Is y = 4 y = 4 a solution? ⓑ Is y = −5 y = −5 a solution?

For the equation t + 6 = t t + 6 = t : ⓐ Is t = −2 t = −2 a solution? ⓑ Is t = 3 t = 3 a solution?

For the equation u + 42 = u u + 42 = u : ⓐ Is u = −6 u = −6 a solution? ⓑ Is u = 7 u = 7 a solution?

In the following exercises, solve.

5 y + 1 = 4 5 y + 1 = 4

7 z + 15 = 6 7 z + 15 = 6

5 x − 6 = 8 5 x − 6 = 8

4 x − 3 = 7 4 x − 3 = 7

2 m − 3 − 5 = 0 2 m − 3 − 5 = 0

2 n − 1 − 3 = 0 2 n − 1 − 3 = 0

6 v − 2 − 10 = 0 6 v − 2 − 10 = 0

4 u + 2 − 6 = 0 4 u + 2 − 6 = 0

5 q + 3 − 4 = 0 5 q + 3 − 4 = 0

4 m + 2 + 2 = 6 4 m + 2 + 2 = 6

6 n + 1 + 4 = 8 6 n + 1 + 4 = 8

2 u − 3 + 2 = 0 2 u − 3 + 2 = 0

5 v − 2 + 5 = 0 5 v − 2 + 5 = 0

3 z − 5 + 2 = 0 3 z − 5 + 2 = 0

2 m + 1 + 4 = 0 2 m + 1 + 4 = 0

ⓐ u − 3 + 3 = u u − 3 + 3 = u ⓑ x + 1 − x + 1 = 0 x + 1 − x + 1 = 0

ⓐ v − 10 + 10 = v v − 10 + 10 = v ⓑ y + 4 − y + 2 = 0 y + 4 − y + 2 = 0

ⓐ r − 1 − r = −1 r − 1 − r = −1 ⓑ z + 100 − z + 10 = 0 z + 100 − z + 10 = 0

ⓐ s − 8 − s = −8 s − 8 − s = −8 ⓑ w + 25 − w + 5 = 0 w + 25 − w + 5 = 0

3 2 x − 3 − 20 = 7 3 2 x − 3 − 20 = 7

2 5 x + 1 − 8 = 0 2 5 x + 1 − 8 = 0

2 8 r + 1 − 8 = 2 2 8 r + 1 − 8 = 2

3 7 y + 1 − 10 = 8 3 7 y + 1 − 10 = 8

3 u − 2 = 5 u + 1 3 u − 2 = 5 u + 1

4 v + 3 = v − 6 4 v + 3 = v − 6

8 + 2 r = 3 r + 10 8 + 2 r = 3 r + 10

12 c + 6 = 10 − 4 c 12 c + 6 = 10 − 4 c

ⓐ a + 2 = a + 4 a + 2 = a + 4 ⓑ b − 2 + 1 = 3 b + 2 b − 2 + 1 = 3 b + 2

ⓐ r + 6 = r + 8 r + 6 = r + 8 ⓑ s − 3 + 2 = s + 4 s − 3 + 2 = s + 4

ⓐ u + 1 = u + 4 u + 1 = u + 4 ⓑ n − 5 + 4 = 3 n + 7 n − 5 + 4 = 3 n + 7

ⓐ x + 10 = x + 2 x + 10 = x + 2 ⓑ y − 2 + 2 = 2 y + 4 y − 2 + 2 = 2 y + 4

2 y + 4 + 6 = 0 2 y + 4 + 6 = 0

8 u + 1 + 9 = 0 8 u + 1 + 9 = 0

a + 1 = a + 5 a + 1 = a + 5

d − 2 = d − 20 d − 2 = d − 20

6 s + 4 = 8 s − 28 6 s + 4 = 8 s − 28

9 p + 9 = 10 p − 6 9 p + 9 = 10 p − 6

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s = A s = A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula s = A s = A to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity While putting up holiday decorations, Renee dropped a light bulb from the top of a 64 foot tall tree. Use the formula t = h 4 t = h 4 to find how many seconds it took for the light bulb to reach the ground.

Gravity An airplane dropped a flare from a height of 1024 feet above a lake. Use the formula t = h 4 t = h 4 to find how many seconds it took for the flare to reach the water.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula t = h 4 t = h 4 to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 54 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 117 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Study Guides > College Algebra

Using the square root property, a general note: the square root property, how to: given a quadratic equation with an [latex]{x}^{2}[/latex] term but no [latex]x[/latex] term, use the square root property to solve it..

• Isolate the [latex]{x}^{2}[/latex] term on one side of the equal sign.
• Take the square root of both sides of the equation, putting a [latex]\pm [/latex] sign before the expression on the side opposite the squared term.
• Simplify the numbers on the side with the [latex]\pm [/latex] sign.

Example 6: Solving a Simple Quadratic Equation Using the Square Root Property

Example 7: solving a quadratic equation using the square root property, licenses & attributions, cc licensed content, specific attribution.

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Quadratic Equations and Functions

Solve Quadratic Equations Using the Square Root Property

Learning objectives.

By the end of this section, you will be able to:

Before you get started, take this readiness quiz.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x 2 = 9.

But what happens when we have an equation like x 2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13) 2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

Since these equations are all of the form x 2 = k , the square root definition tells us the solutions are the two square roots of k . This leads to the Square Root Property .

If x 2 = k , then

Now we will solve the equation x 2 = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x 2 = 7.

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

• Isolate the quadratic term and make its coefficient one.
• Use Square Root Property.
• Simplify the radical.
• Check the solutions.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .

Solve Quadratic Equations of the Form a ( x − h ) 2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a ( x − h ) 2 = k as well. Notice that the quadratic term, x , in the original form ax 2 = k is replaced with ( x − h ).

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a , then the Square Root Property can be used on ( x − h ) 2 .

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

We will start the solution to the next example by isolating the binomial term.

Sometimes the solutions are complex numbers.

The left sides of the equations in the next two examples do not seem to be of the form a ( x − h ) 2 . But they are perfect square trinomials, so we will factor to put them in the form we need.

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

• Solving Quadratic Equations: The Square Root Property
• Using the Square Root Property to Solve Quadratic Equations

Key Concepts

How to solve a quadratic equation using the square root property.

Practice Makes Perfect

Solve Quadratic Equations of the Form ax 2 = k Using the Square Root Property

In the following exercises, solve each equation.

Mixed Practice

In the following exercises, solve using the Square Root Property.

Writing Exercises

In your own words, explain the Square Root Property.

Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no – I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Module 12: Quadratic Equations and Complex Numbers

Quadratic equations, learning objectives.

• Use the zero product principle to solve quadratic equations that can be factored
• Identify solutions to quadratic equations on a graph
• Use the square root property to solve a quadratic equation
• Complete the square to solve a quadratic equation
• Write a quadratic equation in standard form and identify the values of a , b , and c in a standard form quadratic equation.
• Use the Quadratic Formula to find all real solutions of a quadratic equation, recognize when there are no real solutions
• Solve application problems involving quadratic equations

An equation containing a second-degree polynomial is called a quadratic equation . For example, equations such as [latex]2{x}^{2}+3x - 1=0[/latex] and [latex]{x}^{2}-4=0[/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is by  factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. Note that we will not spend a lot of time explaining how to factor in this section.  You may want to seek help if you don’t feel confident about factoring.

Solving by factoring depends on the zero-product property, which states that if [latex]a\cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], where a and b are real numbers or algebraic expressions.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], where a , b , and c are real numbers, and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

The Zero-Product Property and Quadratic Equations

The zero-product property states

where a and b are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

where a , b , and c are real numbers, and if [latex]a\ne 0[/latex], it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is 1. We have one method of factoring quadratic equations in this form.

Reminder: Given a quadratic equation with the leading coefficient of 1, factor it.

• Find two numbers whose product equals c and whose sum equals b .
• Use those numbers to write two factors of the form [latex]\left(x+k\right)\text{ or }\left(x-k\right)[/latex], where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and [latex]-2[/latex], the factors are [latex]\left(x+1\right)\left(x - 2\right)[/latex].
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Factor and solve the equation: [latex]{x}^{2}+x - 6=0[/latex].

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[/latex].

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex].

If we graph the function [latex]f(x)={x}^{2}+x - 6[/latex], we will get the parabola in the figure below. The solutions to the equation[latex]{x}^{2}+x - 6=0[/latex] are the x- intercepts of the function [latex]f(x)={x}^{2}+x - 6[/latex]. Recall that x-intercepts are where the outputs, or y values are zero, therefore the points (-3,0) and (2,0) represent the places where the parabola crosses the x axis.

In the following video we provide more examples of factoring to solve quadratic equations.

In our next example, we will solve a quadratic equation that is written as a difference of squares.

Solve the difference of squares equation using the zero-product property: [latex]{x}^{2}-9=0[/latex].

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

 Solving Quadratics with a Leading Coefficient of [latex]\ne1[/latex]

Recall that when the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

• With the quadratic in standard form, [latex]a{x}^{2}+bx+c=0[/latex], multiply [latex]a\cdot c[/latex].
• Find two numbers whose product equals [latex]ac[/latex] and whose sum equals [latex]b[/latex].
• Rewrite the equation replacing the [latex]bx[/latex] term with two terms using the numbers found in step 1 as coefficients of x.
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Use grouping to factor and solve the quadratic equation: [latex]4{x}^{2}+15x+9=0[/latex].

First, multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex].

The following video contains another example of solving a quadratic equation using factoring with grouping.

Sometimes, we may be given an equation that does not look like a quadratic at first glance. In our next examples we will solve a cubic polynomial equation where the GCF of each term is x, and can be factored.  The result is a quadratic equation that we can solve.

Solve the equation by factoring: [latex]-3{x}^{3}-5{x}^{2}-2x=0[/latex].

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out [latex]-x[/latex] from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

The solutions are [latex]x=0[/latex], [latex]x=-\frac{2}{3}[/latex], and [latex]x=-1[/latex].

In this last video example, we solve a quadratic equation with a leading coefficient of -1 using a shortcut method of factoring and the zero product principle.

Quadratic equations can be solved in many ways. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this topic, you will use square roots to learn another way to solve quadratic equations—and this method will work with all quadratic equations.

Solve a Quadratic Equation by the Square Root Property

One way to solve the quadratic equation [latex]x^{2}=9[/latex] is to subtract 9 from both sides to get one side equal to 0: [latex]x^{2}-9=0[/latex]. The expression on the left can be factored, it is a difference of squares: [latex]\left(x+3\right)\left(x–3\right)=0[/latex]. Using the zero factor property, you know this means [latex]x+3=0[/latex] or [latex]x–3=0[/latex], so [latex]x=−3[/latex] or 3.

Another property would let you solve that equation more easily is called the square root property.

The Square Root Property

If [latex]x^{2}=a[/latex], then [latex] x=\sqrt{a}[/latex] or [latex] -\sqrt{a}[/latex].

The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of a and the negative square root of a .

A shortcut way to write “[latex] \sqrt{a}[/latex]” or “[latex] -\sqrt{a}[/latex]” is [latex] \pm \sqrt{a}[/latex]. The symbol [latex]\pm[/latex] is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”

Solve using the Square Root Property. [latex]x^{2}=9[/latex]

[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]

[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])

Notice that there is a difference here in solving [latex]x^{2}=9[/latex] and finding [latex] \sqrt{9}[/latex]. For [latex]x^{2}=9[/latex], you are looking for all numbers whose square is 9. For [latex] \sqrt{9}[/latex], you only want the principal (nonnegative) square root. The negative of the principal square root is [latex] -\sqrt{9}[/latex]; both would be [latex] \pm \sqrt{9}[/latex]. Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!

In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating [latex]x^{2}[/latex].

In our first video we will show more examples of using the square root property to solve a quadratic equation.

Solve. [latex]10x^{2}+5=85[/latex]

[latex]10x^{2}+5=85[/latex]

You could now take the square root of both sides, but you would have [latex] \sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.

[latex]10x^{2}=80[/latex]

Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily.

Be sure to simplify the radical if possible.

[latex] \begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]

[latex] x=\pm 2\sqrt{2}[/latex]

Sometimes more than just the x is being squared:

Solve. [latex]\left(x–2\right)^{2}–50=0[/latex]

[latex]\left(x-2\right)^{2}-50=0[/latex]

Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.

[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]

To isolate x on the left, you need to add 2 to both sides.

[latex] \begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]

[latex] x=2\pm 5\sqrt{2}[/latex]

In the next video you will see more examples of using square roots to solve quadratic equations.

Solve a Quadratic Equation by Completing the Square

Not all quadratic equations can be factored or solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property.

First, let’s make sure we can recognize a perfect square trinomial and how factor it.

Factor [latex]9x^{2}–24x+16[/latex].

First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares.

[latex]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex]

Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.

[latex]24x=2\left(3x\right)\left(4\right)[/latex]

A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex].

In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex].

[latex]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex]

If this were an equation, we could solve using either the square root property or the zero product property. If you don’t start with a perfect square trinomial, you can complete the square to make what you have into one.

To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

Steps for Completing The Square

We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.

• Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right side of the equal sign. [latex]{x}^{2}+4x=-1[/latex]
• Multiply the b term by [latex]\frac{1}{2}[/latex] and square it. [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
• Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
• The left side of the equation can now be factored as a perfect square. [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
• Use the square root property and solve. [latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
• The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

Solve by completing the square. [latex]x^{2}–12x–4=0[/latex]

First, move the constant term to the right side of the equal sign.Identify b .

[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]

Then, take [latex]\frac{1}{2}[/latex] of the b term and square it. Add [latex] {{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex] {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex].

Add the value to both sides of the equation and simplify.

[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]

Rewrite the left side as a squared binomial.

[latex]\left(x-6\right)^{2}=40[/latex]

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

[latex] x-6=\pm\sqrt{40}[/latex]

Solve for x by adding 6 to both sides. Simplify as needed.

[latex] \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]

[latex] x=6\pm 2\sqrt{10}[/latex]

Solve by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].

First, move the constant term to the right side of the equal sign.

Identify b .[latex]b=-3[/latex] Then, take [latex]\frac{1}{2}[/latex] of the b term and square it.

Add the result to both sides of the equal sign.

Factor the left side as a perfect square and simplify the right side.

Use the square root property and solve.

The solutions are [latex]x=\frac{3}{2}+\frac{\sqrt{29}}{2}[/latex], [latex]x=\frac{3}{2}-\frac{\sqrt{29}}{2}[/latex].

In the next video you will see more examples of how to use completing the square to solve a quadratic equation.

You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.

Solve by completing the square. [latex]x^{2}+16x+17=-47[/latex].

[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]

Add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex] {{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.

[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]

Write the left side as a squared binomial.

[latex]\left(x+8\right)^{2}=0[/latex]

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.

[latex]x+8=0[/latex]

Solve for x.

[latex]x=-8[/latex]

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation [latex]x^{2}+16x+17=−47[/latex] becomes [latex]x^{2}+16x+64=0[/latex]. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).

It can be factored as [latex](x+8)(x+8)=0[/latex], of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.

In our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are rational.

The Quadratic Formula

You can solve any quadratic equation by completing the square —rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[/latex] and then solve for x , you find that [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. This equation is known as the Quadratic Formula.

We can derive the quadratic formula by completing the square. First, assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[/latex] and obtain a positive a . Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:

• First, move the constant term to the right side of the equal sign: [latex]a{x}^{2}+bx=-c[/latex]
• As we want the leading coefficient to equal 1, divide through by a : [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
• Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign: [latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
• Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: [latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
• Now, use the square root property, which gives [latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
• Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus, [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]

This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[/latex].

The form [latex]ax^{2}+bx+c=0[/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it’s vital that you be sure the equation is in this form. If you don’t, you might use the wrong values for a , b , or c , and then the formula will give incorrect solutions.

Solving a Quadratic Equation using the Quadratic Formula

The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, [latex]ax^{2}+bx+c=0[/latex]. To use it, follow these steps.

• Put the equation in standard form first.
• Identify the coefficients, a , b, and c. Be careful to include negative signs if the bx or c terms are subtracted.
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Simplify as much as possible.
• Use the [latex]\pm[/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted .
• Simplify both values to get the possible solutions.

That’s a lot of steps. Let’s try using the Quadratic Formula to solve a relatively simple equation first; then you’ll go back and solve it again using another factoring method.

Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[/latex].

[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\\x^{2}+4x-5=0\,\,\,\\\\a=1,b=4,c=-5\end{array}[/latex]

Note that the subtraction sign means the constant c is negative.

[latex] \begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula. [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\[/latex]

[latex] \begin{array}{l}\\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]

Simplify, being careful to get the signs correct.

[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]

Simplify some more.

[latex] x=\frac{-4\pm \sqrt{36}}{2}[/latex]

Simplify the radical: [latex] \sqrt{36}=6[/latex].

[latex] x=\frac{-4\pm 6}{2}[/latex]

Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.

[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\\\text{or}\\\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]

[latex]x=1\,\,\,\text{or}\,\,\,-5[/latex]

You can check these solutions by substituting [latex]1[/latex] and [latex]−5[/latex] into the original equation.

You get two true statements, so you know that both solutions work: [latex]x=1[/latex] or [latex]-5[/latex]. You’ve solved the equation successfully using the Quadratic Formula!

Watch this video to see an example of how to use the quadratic formula to solve a quadratic equation that has two real, rational solutions.

Sometimes, it may be easier to solve an equation using conventional factoring methods, like finding number pairs that sum to one number (in this example, 4) and that produce a specific product (in this example [latex]−5[/latex]) when multiplied. The power of the Quadratic Formula is that it can be used to solve any quadratic equation, even those where finding number combinations will not work.

In the next video example we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.

Most of the quadratic equations you’ve looked at have two solutions, like the one above. The following example is a little different.

Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[/latex].

[latex]\begin{array}{l}x^{2}-2x=6x-16\\x^{2}-2x-6x+16=0\\x^{2}-8x+16=0\end{array}[/latex]

Identify the coefficients a , b , and c . [latex]x^{2}=1x^{2}[/latex], so [latex]a=1[/latex]. Since [latex]8x[/latex] is subtracted, b is negative. [latex]a=1,b=-8,c=16[/latex]

[latex] \begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]

Substitute the values into the Quadratic Formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]

[latex] x=\frac{8\pm \sqrt{64-64}}{2}[/latex]

Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.

[latex] x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]

[latex]x=4[/latex]

Again, check using the original equation.

[latex]\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\\16-8=24-16\,\,\,\,\,\,\\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

In the next example, we will show that some quadratic equations do not have real solutions.  As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.

Use the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[/latex]

Add x to both sides, and add 3 to both sides to get the quadratic equation in standard form.

[latex]\begin{array}{l}x^{2}+x=-x-3\\x^{2}+2x+3=0\end{array}[/latex]

Identify a, b, c.

[latex]a=1, b=2, c=3[/latex]

Substitute values for a, b, c into the quadratic formula.

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-2\pm \sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\end{array}[/latex]

[latex] x=\frac{-2\pm \sqrt{-8}}{2}[/latex]

Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.

There are no real solutions.

Applying the Quadratic Formula

Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn’t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.

A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called initial velocity ) is [latex]−10[/latex] feet per second. (The negative value means it’s heading toward the ground.)

The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x . [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex] t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex] \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.

The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

In the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.

The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.

Bob made a quilt that is 4 ft [latex]\times[/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.

Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].

(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)

You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are 10 sq ft of fabric for the border, so set the area of border to be 10.

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

[latex]10=18x+4x^{2}[/latex]

Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}-10\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

[latex] x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative.

The width of the border should be 0.5 ft.

Our last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.

Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations, or any other quadratic equation! The Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex], is found by completing the square of the quadratic equation [latex] [/latex]. When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.

You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if [latex]ab=0[/latex], then either [latex]a=0[/latex] or [latex]b=0[/latex], or both a and b are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.

Not all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded.

Completing the square is used to change a binomial of the form [latex]x^{2}+bx[/latex] into a perfect square trinomial [latex] {{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}[/latex], which can be factored to [latex] {{\left( x+\frac{b}{2} \right)}^{2}}[/latex]. When solving quadratic equations by completing the square, be careful to add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex] to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x . With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.

• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Quadratic Formula Application - Time for an Object to Hit the Ground. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/RcVeuJhcuL0 . License : CC BY: Attribution
• Quadratic Formula Application - Determine the Width of a Border. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/Zxe-SdwutxA . License : CC BY: Attribution
• College Algebra. Authored by : Abramson, Jay, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at : http://cnx.org/contents/[email protected]:1/Preface
• Ex: Solve a Quadratic Equation Using Factor By Grouping. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/04zEXaOiO4U . License : CC BY: Attribution
• Ex: Factor and Solve Quadratic Equation - Trinomial a = -1. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/nZYfgHygXis . License : CC BY: Attribution
• Unit 12: Factoring, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
• Ex 1: Solving Quadratic Equations Using Square Roots. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/Fj-BP7uaWrI . License : CC BY: Attribution
• Ex 2: Solving Quadratic Equations Using Square Roots. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/4H5qZ_-8YM4 . License : Public Domain: No Known Copyright
• Ex 1: Completing the Square - Real Rational Solutions. Authored by : James Sousa (Mathispower4u.com) . License : Public Domain: No Known Copyright
• Ex 2: Completing the Square - Real Irrational Solutions. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/IjCjbtrPWHM . License : CC BY: Attribution
• Ex2: Quadratic Formula - Two Real Irrational Solutions. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/tF0muV86dr0 . License : CC BY: Attribution
• Ex: Quadratic Formula - Two Real Rational Solutions. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/xtwO-n8lRPw . License : CC BY: Attribution

Algebra Examples

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