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9.6: Solve Applications of Quadratic Equations

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Learning Objectives

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

• Solve applications modeled by quadratic equations

Before you get started, take this readiness quiz.

• The sum of two consecutive odd numbers is \(−100\). Find the numbers. If you missed this problem, review Example 2.18.
• Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{x^{2}-1}\). If you missed this problem, review Example 7.35.
• Find the length of the hypotenuse of a right triangle with legs \(5\) inches and \(12\) inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

• Square Root Property
• Completing the Square
• Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy

• Read the problem. Make sure all the words and ideas are understood.
• Identify what we are looking for.
• Name what we are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Solve the equation using algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is \(2\) more than the number preceding it. If we call the first one \(n\), then the next one is \(n+2\). The next one would be \(n+2+2\) or \(n+4\). This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

\(\begin{array}{cl}{}&{\text{Consecutive even integers}}\\{}& {64,66,68}\\ {n} & {1^{\text { st }} \text { even integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive even integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive even integer }}\end{array}\)

\(\begin{array}{cl}{}&{\text{Consecutive odd integers}}\\{}& {77,79,81}\\ {n} & {1^{\text { st }} \text { odd integer }} \\ {n+2} & {2^{\text { nd }} \text { consecutive odd integer }} \\ {n+4} & {3^{\text { rd }} \text { consecutive odd integer }}\end{array}\)

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example \(\PageIndex{1}\)

The product of two consecutive odd integers is \(195\). Find the integers.

Step 1 : Read the problem

Step 2 : Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3 : Name what we are looking for.

Let \(n=\) the first odd integer.

\(n+2=\) the next odd integer.

Step 4 : Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is \(195\).” The product of the first odd integer and the second odd integer is \(195\).

Translate into an equation.

\(n(n+2)=195\)

Step 5 : Solve the equation. Distribute.

\(n^{2}+2 n=195\)

Write the equation in standard form.

\(n^{2}+2 n-195=0\)

\((n+15)(n-13)=0\)

Use the Zero Product Property.

\(n+15=0 \quad n-13=0\)

Solve each equation.

\(n=-15, \quad n=13\)

There are two values of \(n\) that are solutions. This will give us two pairs of consecutive odd integers for our solution.

\(\begin{array}{cc}{\text { First odd integer } n=13} & {\text { First odd integer } n=-15} \\ {\text { next odd integer } n+2} & {\text { next odd integer } n+2} \\ {13+2} & {-15+2} \\ {15} & {-13}\end{array}\)

Step 6 : Check the answer.

Do these pairs work? Are they consecutive odd integers?

\(\begin{aligned} 13,15 & \text { yes } \\-13,-15 & \text { yes } \end{aligned}\)

Is their product \(195\)?

\(\begin{aligned} 13 \cdot 15 &=195 &\text{yes} \\-13(-15) &=195 & \text { yes } \end{aligned}\)

Step 7 : Answer the question.

Two consecutive odd integers whose product is \(195\) are \(13,15\) and \(-13,-15\).

Exercise \(\PageIndex{1}\)

The product of two consecutive odd integers is \(99\). Find the integers.

The two consecutive odd integers whose product is \(99\) are \(9, 11\), and \(−9, −11\).

Exercise \(\PageIndex{2}\)

The product of two consecutive even integers is \(168\). Find the integers.

The two consecutive even integers whose product is \(128\) are \(12, 14\) and \(−12, −14\).

We will use the formula for the area of a triangle to solve the next example.

Definition \(\PageIndex{1}\)

Area of a Triangle

For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2} b h\).

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example \(\PageIndex{2}\)

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of \(120\) square feet and the architect wants the base to be \(4\) feet more than twice the height. Find the base and height of the window.

Exercise \(\PageIndex{3}\)

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of \(456\) square inches.

The height of the triangle is \(12\) inches and the base is \(76\) inches.

Exercise \(\PageIndex{4}\)

If a triangle that has an area of \(110\) square feet has a base that is two feet less than twice the height, what is the length of its base and height?

The height of the triangle is \(11\) feet and the base is \(20\) feet.

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Definition \(\PageIndex{2}\)

Area of a Rectangle

For a rectangle with length, \(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

Example \(\PageIndex{3}\)

Mike wants to put \(150\) square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than \(3\) times the width. Find the length and width. Round to the nearest tenth of a foot.

Exercise \(\PageIndex{5}\)

The length of a \(200\) square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

The length of the garden is approximately \(18\) feet and the width \(11\) feet.

Exercise \(\PageIndex{6}\)

A rectangular tablecloth has an area of \(80\) square feet. The width is \(5\) feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot?

The length of the tablecloth is approximately \(11.8\) feet and the width \(6.8\) feet.

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Definition \(\PageIndex{3}\)

Pythagorean Theorem

• In any right triangle, where \(a\) and \(b\) are the lengths of the legs, and \(c\) is the length of the hypotenuse, \(a^{2}+b^{2}=c^{2}\).

Example \(\PageIndex{4}\)

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two \(10\)-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Exercise \(\PageIndex{7}\)

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is \(20\) feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

The length of the flag pole’s shadow is approximately \(6.3\) feet and the height of the flag pole is \(18.9\) feet.

Exercise \(\PageIndex{8}\)

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The distance between the opposite corners is approximately \(7.2\) feet.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, \(v_{0}\), propels the object up until gravity causes the object to fall back down.

Definition \(\PageIndex{4}\)

The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula

\(h=-16 t^{2}+v_{0} t\)

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example \(\PageIndex{5}\)

A firework is shot upwards with initial velocity \(130\) feet per second. How many seconds will it take to reach a height of \(260\) feet? Round to the nearest tenth of a second.

Exercise \(\PageIndex{9}\)

An arrow is shot from the ground into the air at an initial speed of \(108\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the arrow will be \(180\) feet from the ground. Round the nearest tenth.

The arrow will reach \(180\) feet on its way up after \(3\) seconds and again on its way down after approximately \(3.8\) seconds.

Exercise \(\PageIndex{10}\)

A man throws a ball into the air with a velocity of \(96\) ft/s. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the height of the ball will be \(48\) feet. Round to the nearest tenth.

The ball will reach \(48\) feet on its way up after approximately \(.6\) second and again on its way down after approximately \(5.4\) seconds.

We have solved uniform motion problems using the formula \(D=rt\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula \(D=rt\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\frac{D}{r}\).

Some uniform motion problems are also modeled by quadratic equations.

Example \(\PageIndex{6}\)

Professor Smith just returned from a conference that was \(2,000\) miles east of his home. His total time in the airplane for the round trip was \(9\) hours. If the plane was flying at a rate of \(450\) miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

We are looking for the speed of the jet stream. Let \(r=\) the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is \(450 + r\).

When the plane flies against the wind, the wind decreases its speed and the rate is \(450 − r\).

The speed of the jet stream was \(50\) mph.

Exercise \(\PageIndex{11}\)

MaryAnne just returned from a visit with her grandchildren back east. The trip was \(2400\) miles from her home and her total time in the airplane for the round trip was \(10\) hours. If the plane was flying at a rate of \(500\) miles per hour, what was the speed of the jet stream?

The speed of the jet stream was \(100\) mph.

Exercise \(\PageIndex{12}\)

Gerry just returned from a cross country trip. The trip was \(3000\) miles from his home and his total time in the airplane for the round trip was \(11\) hours. If the plane was flying at a rate of \(550\) miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example \(\PageIndex{7}\)

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(12\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(8\) hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Exercise \(\PageIndex{13}\)

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes \(6\) hours more than Press #2 to do the job and when both presses are running they can print the job in \(4\) hours. How long does it take for each press to print the job alone?

Press #1 would take \(12\) hours, and Press #2 would take \(6\) hours to do the job alone.

Exercise \(\PageIndex{14}\)

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes \(3\) hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in \(2\) hours. How long does it take for each hose to fill the hot tub?

The red hose take \(6\) hours and the green hose take \(3\) hours alone.

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

• Word Problems Involving Quadratic Equations
• Quadratic Equation Word Problems
• Applying the Quadratic Formula

Key Concepts

• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
• Solve the equation using good algebra techniques.
• For a triangle with base, \(b\), and height, \(h\), the area, \(A\), is given by the formula \(A=\frac{1}{2}bh\).

• For a rectangle with length,\(L\), and width, \(W\), the area, \(A\), is given by the formula \(A=LW\).

• The height in feet, \(h\), of an object shot upwards into the air with initial velocity, \(v_{0}\), after \(t\) seconds is given by the formula \(h=-16 t^{2}+v_{0} t\).

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Solving quadratic equation

Here you will learn about solving quadratic equations and how to do it using a graph, factoring the equation or using the quadratic formula.

Students will first learn about solving quadratic equations as part of algebra in high school.

What is solving quadratic equations?

Solving quadratic equations is finding the roots or the x- intercepts of the parabola formed by a quadratic equation.

For example,

Solve 2x^2+7x-4 by using its graph below.

The parabola passes through the x- axis as (-4,0) and \left(\cfrac{1}{2}, 0\right).

The solutions are x=-4 and x=\cfrac{1}{2}.

Solve 2x^2+7x-4=0 by factoring.

Find where each value being multiplied will be equal to 0.

2x-1=0, so x=\cfrac{1}{2}

x+4=0 so x=-4

Solve 2x^2+7x-4=0 using the quadratic formula.

The quadratic formula uses the values from the constant terms a, b and c when quadratic equations are in the general form ax^2+b x+c=0.

\begin{aligned} & x=\cfrac{-b+\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7+\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7+\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7+9}{4} \\\\ & x=\cfrac{2}{4} \\\\ & x=\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-b-\sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-7-\sqrt{7^2-4 \times 2 \times(-4)}}{2 \times 2} \\\\ & x=\cfrac{-7-\sqrt{49-(-32)}}{4} \\\\ & x=\cfrac{-7-9}{4} \\\\ & x=\cfrac{-16}{4} \\\\ & x=-4 \end{aligned}

Common Core State Standards

How does this relate to high school math?

• Algebra – Reasoning with Equations and Inequalities (HSA.REI.B.4b) Solve quadratic equations in one variable. Solve quadratic equations by inspection ( e.g., for x^2=49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a \pm bi for real numbers a and b.

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How to solve a quadratic equation by graphing

In order to solve quadratic equations by graphing:

• Graph the quadratic equation.
• Identify the \textbf{x}- intercepts.

Solving quadratic equation examples

Example 1: solve a quadratic equation by looking at its graph.

Solve the following equation by using a graph: x^2-x-6.

2 Identify the \textbf{x}- intercepts.

The parabola passes through the x- axis as (-2,0) and (3,0)

The solutions are x=-2 and x=3.

Example 2: solve a quadratic equation by looking at its graph

Solve the following equation by using a graph: 2x^2-3x+1.125.

The parabola passes through the x- axis as (0.75,0). Since it crosses at the vertex of the parabola, there is only 1 solution.

The solution is x=0.75.

See also : Quadratic graphs

How to solve a quadratic equation by factoring

In order to solve quadratic equations by factoring:

• Make sure that the equation is equal to \bf{0}.
• Fully factor the quadratic equation .
• Set each expression equal to \bf{0}.
• Solve for \textbf{x}.

Example 3: solve a quadratic equation by factoring with coefficient a=1

Solve x^{2}-8x+15=0 by factoring.

(x-3)(x-5)=0

x-3=0 \quad x-5=0

The opposite of -3 is +3, so add 3 to both sides of the equation.

The opposite of -5 is +5, so add 5 to both sides of the equation.

You can see the real roots of the quadratic equation are where the quadratic graph crosses the x- axis.

Example 4: solve a quadratic equation by factoring with coefficient a>1

Solve 2x^{2}+5x+3=0 by factoring.

(2x+3)(x+1)=0

2 x+3=0 \quad x+1=0

The opposite of +3 is -3, so subtract 3 to both sides of the equation.

The opposite of \times 2 is \div 2, so divide by 2 on both sides of the equation.

The opposite of +1 is -1, so -1 to both sides of the equation.

How to solve quadratic equations with the quadratic formula

In order to solve quadratic equations with the quadratic formula:

• Identify the value of \bf{\textbf{a}, \textbf{b}} and \textbf{c} in a quadratic equation.

Substitute these values into the quadratic formula.

• Solve the equation with a \bf{+}, and then with a \bf{-}.

Example 5: equation of the form ax²+b x+c=0 where a = 1

Solve x^2-2 x-15=0 using the quadratic formula.

\begin{aligned} & x=\cfrac{-(-2)+\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2+\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2+8}{2} \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-2)-\sqrt{(-2)^2-4(1)(-15)}}{2(1)} \\\\ & x=\cfrac{2-\sqrt{4-(-60)}}{2} \\\\ & x=\cfrac{2-8}{2} \\\\ & x=-3 \end{aligned}

You can see the roots of the quadratic equation are where the quadratic graph crosses the x- axis.

You can also check that the solutions are correct by substituting them into the original equation.

Example 6: solutions of the quadratic equation with a<1

Solve -6 x^2+2 x-x=-2 using the quadratic formula.

First, simplify the equation so that it is in the form ax^2+b x+c=0, which leaves a quadratic expression on the left side of the equation and 0 on the right side of the equation.

-6 x^2+x=-2 \hspace{0.5cm} *Combine the like terms (2x-x)

-6 x^2+x=-2 \hspace{0.5cm} *Add 2 to both sides so that the equation equals 0

\hspace{0.8cm} +2 \hspace{0.4cm}+2

-6 x^2+x+2=0

\begin{aligned} & x=\cfrac{-(1)+\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1+\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1+7}{-12} \\\\ & x=-\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-(1)-\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1-\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1-7}{-12} \\\\ & x=\cfrac{2}{3} \end{aligned}

Teaching tips for solving quadratic equations

• Let students use a graphing calculator to explore how changing the values for a, b and c in a quadratic equation changes the parabola.
• When students start using the quadratic formula, still encourage them to explain the meaning of their solutions. This way they do not forget what exactly they are solving for. One way to do this is to incorporate many real world examples of the quadratic equation that require students to explain the roots within the context of the problem.

Easy mistakes to make

• Thinking the order of the factored equation matters When you multiply two values the order doesn’t matter. For example, 2\times 3=3\times 2 It is the same with quadratics. (x-6)(x+4) means (x-6)\times (x+4) So, (x-6)(x+4)=0 is the same as (x+4)(x-6)=0.
• Forgetting to solve after factoring Don’t forget to set the factored expression equal to zero and solve it. Always check you have answered the question.

• Thinking an equation that cannot be factored has no solutions If a quadratic equation cannot be factored, you can still solve it by using the quadratic formula. To work out the number of real solutions a quadratic equation has you can use the discriminant.
• Not converting to the form \bf{\textbf{ax}^{2}+\textbf{b x}+\textbf{c}=0} before trying to factor While it is possible to factor a quadratic equation without it in the standard form, this can be challenging. It is recommended to use the general form shown.

Related quadratic equation lessons

• Quadratic formula
• Solving quadratic equations by graphing
• Factoring quadratic equations

Solving quadratic equations questions (by factoring)

1) Solve -x^2+2x by using its graph below.

The parabola passes through the x- axis as (0,0) and (2,0).

The solutions are x = 0 and x = 2.

2) Solve x^2+10 x+9 by using its graph below.

The parabola passes through the x- axis as (-1,0) and (-9,0).

The solutions are x = -1 and x = -9.

3) Solve {x}^2+5x+6=0 by factoring.

{x}^2+5x+6=0 can be factored as (x+2)(x+3).

Setting each factor equal to zero and solving leads to the solutions.

\begin{aligned} & x+2=0 \\\\ & x=-2 \end{aligned} \quad \begin{aligned} & x+3=0 \\\\ & x=-3 \end{aligned}

4) Solve {x}^2-x-20=0 by factoring.

x^2-x-20 can be factored as (x-5)(x+4).

\begin{aligned} & x-5=0 \\\\ & x=5 \end{aligned} \quad \begin{aligned} & x+4=0 \\\\ & x=-4 \end{aligned}

5) Solve 2{x}^2+3x-9=0 using the quadratic equation.

x=-\cfrac{3}{2}=-1.54 and x= -3

x=\cfrac{3}{2}=1.5 and x=3

x=-\cfrac{3}{2}=-1.5 and x=3

x=\cfrac{3}{2}=1.5 and x=-3

Identify a, b and c.

Solve the equation with a +, and then with a -.

\begin{aligned} & x=\cfrac{-3+\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3+\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3+9}{4} \\\\ & x=\cfrac{3}{2}=1.5 \end{aligned} \quad \begin{aligned} & x=\cfrac{-3-\sqrt{(3)^2-4(2)(-9)}}{2(2)} \\\\ & x=\cfrac{-3-\sqrt{9-(-72)}}{4} \\\\ & x=\cfrac{-3-9}{4} \\\\ & x=-3 \end{aligned}

*Divide the numerator by the denominator to convert \cfrac{3}{2} to decimal form.

6) Solve 3{x}^2-9x+6=0 using the quadratic equation.

\begin{aligned} & x=\cfrac{-(-9)+\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9+\sqrt{81-72}}{6} \\\\ & x=\cfrac{9+3}{6} \\\\ & x=2 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-9)-\sqrt{(-9)^2-4(3)(6)}}{2(3)} \\\\ & x=\cfrac{9-\sqrt{81-72}}{6} \\\\ & x=\cfrac{9-3}{6} \\\\ & x=1 \end{aligned}

Solving quadratic equation FAQs

It is a polynomial equation whose highest variable is to the second-degree ( exponent of 2).

The standard form is ax^2+b x+c=0 or f(x)=a x^2+b x+c. The coefficients a, b and c can be whole numbers, integers, fractions, decimals or any other real number where a ≠ 0.

It is the value below the radical in the quadratic formula. If the discriminant b^2-4ac is positive, there are two solutions. If it is 0, there is 1 solution. If it is negative, there are no real solutions.

The next lessons are

• Angles in parallel lines

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Quadratic Equations | Solved Problems and Practice Questions

In this article we cover quadratic equations – definitions, formats, solved problems and sample questions for practice.

A quadratic equation is a polynomial whose highest power is the square of a variable (x 2 , y 2 etc.)

Definitions

A monomial is an algebraic expression with only one term in it.

Example: x 3 , 2x, y 2 , 3xyz etc.

A polynomial is an algebraic expression with more than one term in it.

Alternatively it can be stated as –

A polynomial is formed by adding/subtracting multiple monomials.

Example: x 3 +2y 2 +6x+10, 3x 2 +2x-1, 7y-2 etc.

A polynomial that contains two terms is called a binomial expression .

A polynomial that contains three terms is called a trinomial expression .

A standard quadratic equation looks like this:

ax 2 +bx+c = 0

Where a, b, c are numbers and a≥1.

a, b are called the coefficients of x 2 and x respectively and c is called the constant.

The following are examples of some quadratic equations:

1) x 2 +5x+6 = 0 where a=1, b=5 and c=6.

2) x 2 +2x-3 = 0 where a=1, b=2 and c= -3

3) 3x 2 +2x = 1

→ 3x 2 +2x-1 = 0 where a=3, b=2 and c= -1

4) 9x 2 = 4

→ 9x 2 -4 = 0 where a=9, b=0 and c= -4

For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.

For a quadratic equation ax 2 +bx+c = 0,

the sum of its roots = –b/a and the product of its roots = c/a.

A quadratic equation may be expressed as a product of two binomials.

For example, consider the following equation

x 2 -(a+b)x+ab = 0

x 2 -ax-bx+ab = 0

x(x-a)-b(x-a) = 0

(x-a)(x-b) = 0

x-a = 0 or x-b = 0 x = a or x=b

Here, a and b are called the roots of the given quadratic equation.

Now, let’s calculate the roots of an equation x 2 +5x+6 = 0.

We have to take two numbers adding which we get 5 and multiplying which we get 6. They are 2 and 3.

Let us express the middle term as an addition of 2x and 3x.

→ x 2 +2x+3x+6 = 0

→ x(x+2)+3(x+2) = 0

→ (x+2)(x+3) = 0

→ x+2 = 0       or         x+3 = 0

→ x = -2          or         x = -3

This method is called factoring .

We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.

Sum of the roots for the equation x 2 +5x+6 = 0 is -5 and the product of the roots is 6.

The roots of this equation -2 and -3 when added give -5 and when multiplied give 6.

Solved examples of Quadratic equations

Let us solve some more examples using this method. Problem 1: Solve for x: x 2 -3x-10 = 0

Let us express -3x as a sum of -5x and +2x.

→ x 2 -5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0        or         x+2 = 0

→ x = 5           or         x = -2 Problem 2: Solve for x: x 2 -18x+45 = 0

The numbers which add up to -18 and give +45 when multiplied are -15 and -3.

Rewriting the equation,

→ x 2 -15x-3x+45 = 0

→ x(x-15)-3(x-15) = 0

→ (x-15) (x-3) = 0

→ x-15 = 0      or         x-3 = 0

→ x = 15         or         x = 3

Till now, the coefficient of x 2 was 1. Let us see how to solve the equations where the coefficient of x 2 is greater than 1. Problem 3: Solve for x: 3x 2 +2x =1

Rewriting our equation, we get 3x 2 +2x-1= 0

Here, the coefficient of x 2 is 3. In these cases, we multiply the constant c with the coefficient of x 2 . Therefore, the product of the numbers we choose should be equal to -3 (-1*3).

Expressing 2x as a sum of +3x and –x

→ 3x 2 +3x-x-1 = 0

→ 3x(x+1)-1(x+1) = 0

→ (3x-1)(x+1) = 0

→ 3x-1 = 0      or         x+1 = 0

→ x = 1/3        or         x = -1 Problem 4: Solve for x: 11x 2 +18x+7 = 0

In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.

This can be done by expressing 18x as the sum of 11x and 7x.

→ 11x 2 +11x+7x+7 = 0

→ 11x(x+1) +7(x+1) = 0

→ (x+1)(11x+7) = 0

→ x+1 = 0       or         11x+7 = 0

→ x = -1          or         x = -7/11.

The factoring method is an easy way of finding the roots. But this method can be applied only to equations that can be factored.

For example, consider the equation x 2 +2x-6=0.

If we take +3 and -2, multiplying them gives -6 but adding them doesn’t give +2. Hence this quadratic equation cannot be factored.

For this kind of equations, we apply the quadratic formula to find the roots.

The quadratic formula to find the roots,

x = [-b ± √(b 2 -4ac)] / 2a

Now, let us find the roots of the equation above.

x 2 +2x-6 = 0

Here, a = 1, b=2 and c= -6.

Substituting these values in the formula,

x = [-2 ± √(4 – (4*1*-6))] / 2*1

→ x = [-2 ± √(4+24)] / 2

→ x = [-2 ± √28] / 2

When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.

→ x = [-2 ± √(4*7)] / 2

→ x = [-2 ± 2√7] / 2

→ x = 2[ -1 ± √7] / 2

→ x = -1 ± √7

Hence, √7-1 and -√7-1 are the roots of this equation.

Let us consider another example.

Solve for x: x 2 = 24 – 10x

Rewriting the equation into the standard quadratic form,

x 2 +10x-24 = 0

What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.

So this can be solved by the factoring method. But let’s solve it using the new method, applying the quadratic formula.

Here, a = 1, b = 10 and c = -24.

x = [-10 ± √(100 – 4*1*-24)] / 2*1

x = [-10 ± √(100-(-96))] / 2

x = [-10 ± √196] / 2

x = [-10 ± 14] / 2

x = 2 or x= -12 are the roots.

Discriminant

For an equation ax 2 +bx+c = 0, b 2 -4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation.

If b 2 -4ac > 0, the roots are real and distinct.

If b 2 -4ac = 0, the roots are real and equal.

If b 2 -4ac < 0, the roots are not real (they are complex).

Consider the following example:

Problem: Find the nature of roots for the equation x 2 +x+12 = 0.

b 2 -4ac = -47 for this equation. So it has complex roots. Let us verify this.

→ [ -1±√(1-48)] / 2(1)

→ [-1±√-47] / 2

√-47 is usually written as i √47 indicating it’s an imaginary number.

Hence verified.  

Quadratic Equations Quiz: Solve the following

Problem 1: Click here

A. x = 14 or x = 4 B. x = 8 or x = 7 C. x = 28 or x = 2 D. All of the above

Answer 1: Click here

Explanation

Only 8 and 7 satisfy the conditions of adding up to 15 and giving a product of 56.

Problem 2: Click here

Find x if 2x 2 +7x+4 = 0

A. -7 ± √17 / 4 B. -7 ± √7 / 4 C. [-7 ± √17] / 4 D. [-7 ± √17] / 2

Answer 2: Click here

Explanation :

Applying the quadratic formula and substituting a=2, b=7 and c=4, we get the answer as C.

Problem 3: Click here

For what value of k does the equation x 2 -12x+k = 0 have real and equal roots? A. 6 B. 35 C. 12 D. 36

Answer 3: Click here

b 2 -4ac = 0 for the equation to have real and equal roots. 144-4k = 0 → k = 36

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34 thoughts on “Quadratic Equations | Solved Problems and Practice Questions”

36(x)^2-24(2)^1/2 x+8 = 0 solve it

Very good question

Roots: [x – {2(2^1/2)}/3] [x – {2(2^1/2)}/3]

Credit: Pro. Po-Shen Loh

36x^2 – 24(√2)x + 8=0

x =24√2±[√(1152 – 4*36*8)]/2*36 =24√2±[√1152-1152]/72 =24√2±0/72 so, x =24√2/72 you can simplify ahead by dividing further x=√2 / 3

Yo how on earth is 0/72 = 72. It is zero. What nonsense. So ans is 24√2

There should be a bracket [(24 √(2)±0)/72] and hence the answer is (√(2)/3).

do not say something is nonsense..even if you have second opinions about them there are some students who learn from this site…please be mindful of your words

If p and q are the root of the equation ax2+bx+c=0,than the equation whose roots are p+1/q and q+1 ?

acx^2 + b(a+c)x + a^2 + b^2 +2ac=0

is it p + (1/q) or (p+1)/q?

It will be , X^2-{p+(1/q) +q+1}x+pq+P+(1/q) +1

Find the valu of x and y. Root x +y=7 and x+ root y =11 were x=2

u may use elimination method to solve for x any y

2x-10x+9=0 Here, a=2,b= -10,c=9 X=[(-b)±√ (b² – 4ac)]/2a = [-(-10)± √{(-10)² – 4×2×9}]/2×2 ={10±√(100 – 72)}/4 ={10± √28}/4 ={10±2√7}/4 =(10+2√7)/4 or (10 – 2√7)/4 ={2(5+√7)}/4 or {2(5 – √7)}/4 X=(5+√7)/2 or (5 – √7)/2

Solve it x2-20x+a2-b2=0

show that x²+2x=(2a+2b+1)(2a+2b-1) are the root of the equation are integers

If a+b is a root of square of X+ax+b equals 0 then what is the maximum value of square of b?

In 3/(x+2)(2x-1) = a/(2x+1) + b/(x+2), what is the value of a and b

The sum of the roots of a quadratic equation is 5/2 and the product of its root is 4. The quadratic equation is_______

Equation is 2x^2-5x+4=0 with roots (5+i7)/4 & (5-7i)/4

The sum of a quadratic equation is k+2 and the product if its roots is 6-k^2.the quadratic equation is ___________?

what is the quadratic equation 15-2x-x=0

Using a graph of the equation y=2x³+5x²-x-6 for -4≤x≥2. Use 2cm to represent 1 unit on the x-axis and 1cm to represent 5 units on the y-axis. Then solve the equations: 1) 2x³+5x²+x-4 2) 2x³ +5x²-x+2

Please tell solution of x^2-4ax+4a^2-9=0

find the limit between which k must lie in order that kx square -6x+4 over4x square -6x+x may be capable of all values when x is real

p+1/q =0 q+1=0 q=-1 p+1/-1=0 then p=1 to find the equation we use (x-x1) (x-x2)=0 , (x-(-1)) (x-1)=0, (x+1) (x-1)=0 the quadratic equation x^2 -x+x -1=0, x^2-1=0

Thanks but I got a problem with this number Find the possible values of k if the equation 2kx^2-8x+1=2k(x-2) has equal roots.

In the question find for balie of x, X² + 10X – 24, X = +2 or -12 If x = -2 or 12, The equation would be X² – 10X – 24

Solve 2ײ-11×+14

It’s a good platform and I’ve learnt a lot. However, i request that you include question and solutions for questions like” show that the equation X2+(2-k)x+k=3 has equal roots for all k values?” I would really appreciate for the help. Thanks you God bless you!!!

Solve the quadratic equation by finding I.F ( x-x^2y)dy + (y+xy^2)dx=0

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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

• First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
• Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

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• Projectile Problems

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• Maths Questions

Quadratic Formula Questions

Quadratic formula questions are given here to help students of Classes 10 and 11 in solving quadratic equations, whose roots are not real numbers. As we know, finding the roots of quadratic equations is simple using the factorisation technique, but there are some cases where we cannot use the approach. In such cases, we must use a technique that helps us find the roots of quadratics effortlessly. That is the quadratic formula.

What is the Quadratic formula?

In mathematics, the quadratic formula is one of the effective techniques for solving quadratic equations, and finding the roots of quadratic equations. The quadratic formula for calculating the roots of a quadratic equation ax 2 + bx + c = 0 is given by:

x = [-b ± √(b 2 – 4ac)]/2a

Here, D = b 2 – 4ac is called the discriminant of the quadratic equation, and based on this, we can define the nature of the roots of a quadratic equation.

• If D < 0, the roots are not real, i.e., imaginary.
• If D = 0, the roots are real and equal.
• If D > 0, the roots are real and unequal.

Learn more about the Quadratic formula and its derivation here.

Also, check other methods of finding the roots of quadratic equations given below.

• Factorisation
• Completing the square

However, the quadratic formula is used to find the roots of a quadratic equation when the above two methods are not sufficient, i.e., to find the imaginary roots.

Quadratic Formula Questions and Answers

1. Using the quadratic formula, find the roots of the quadratic equation 2x 2 – 7x + 6 = 0.

2x 2 – 7x + 6 = 0

When compared with the standard form, we have;

a = 2, b = -7 and c = 6

D = b 2 – 4ac

= (-7) 2 – 4(2)(6)

= 49 – 48

That means the roots are real and unequal.

Using the quadratic formula,

= [-(-7) ± √1]/ 2(2)

= (7 ± 1)/4

x = (7 + 1)/4, x = (7 – 1)/4

x = 8/4, x = 6/4

x = 2, x = 3/2

Therefore, 2 and 3/2 are the roots of the given equation.

2. Find whether the equation 5x 2 – 2x – 10 = 0 has real roots. If real roots exist, find them.

5x 2 – 2x – 10 = 0

Here, a = 5, b = -2, and c = -10

Let us calculate the discriminant value.

= (-2) 2 – 4(5)(-10)

= 204 > 0

That means the given quadratic equation has two real and unequal roots.

Now, using the quadratic formula, we have;

= [-(-2) ± √204]/ 2(5)

= (2 ± √(4 × 51))/10

x = (2 + 2√51)/10, x = (2 – 2√51)/10

x = 2(1 + √51)/10, x = 2(1 – √51)/10

x = (1 + √51)/5 , x = (1 – √51)/5

Therefore, the real roots of the given quadratic equation are (1 + √51)/5 and (1 – √51)/5.

3. Calculate the roots of the quadratic equation x – 1/x = 3, x ≠ 0, using the quadratic formula.

Given the quadratic equation is:

x – 1/x = 3

(x 2 – 1)/x = 3

x 2 – 1 = 3x

x 2 – 3x – 1 = 0

Here, a = 1, b = -3 and c = -1

= (-3) 2 – 4(1)(-1)

= 13 > 0

Thus, the roots of the given quadratic equation are real and unequal.

= [-(-3) ± √13]/ 2(1)

= (3 ± √13)/2

x = (3 + √13)/2, x = (3 – √13)/2

Hence, (3 + √13)/2 and (3 – √13)/2 are the roots of the given quadratic equation.

4. What are the roots of the quadratic equation 2x 2 – √5x + 1 = 0?

2x 2 – √5x + 1 = 0

Here, a = 2, b = -√5 and c = 1

= (-√5) 2 – 4(2)(1)

= 5 – 8

= -3 < 0

That means the roots are imaginary.

Using the quadratic formula, we have;

= [-(-√5) ± √(-3)]/ 2(2)

= (√5 ± i√3)/4

x = (√5 + i√3)/4, (√5 – i√3)/4

Hence, (√5 + i√3)/4 and (√5 – i√3)/4 are the roots of the quadratic equation 2x 2 – √5x + 1 = 0.

5. Find the roots of the quadratic equation 4x 2 + 4√3x + 3 = 0.

4x 2 + 4√3x + 3 = 0

Here, a = 4, b = 4√3 and c = 3

b 2 – 4ac = (4√3) 2 – 4(4)(3)

= 48 – 48

That means the roots of the given quadratic equation are real and equal.

By using the quadratic formula, we get,

x = (-4√3 ± √0)/8

x = (-4√3)/8

Therefore, -√3/2 and -√3/2 are the roots of the given equation.

6. What will be the roots of the quadratic equation 10x 2 − 9x + 6 = 0?

10x 2 − 9x + 6 = 0

Here, a = 10, b = -9 and c = 6.

b 2 – 4ac = (-9) 2 – 4(10)(6)

= 81 – 240

x = [-(-9) ± √(b 2 – 4ac)]/2a

= [9 ± √(-159)]/ 2(10)

= (9 ± i√159)/20

x = (9 + i√159)/20, (9 – i√159)/20

Hence, (9 + i√159)/20 and (9 – i√159)/20 are the roots of the quadratic equation 10x 2 − 9x + 6 = 0.

7. Solve the following equation using the quadratic formula.

4x 2 – 1 = -8x

4x 2 + 8x – 1 = 0

Here, a = 4, b = 8 and c = -1

Now, b 2 – 4ac = (8) 2 – 4(4)(-1)

= 80 > 0

That means the given quadratic equation contains two real and unequal roots.

Now, by using the quadratic formula, we have;

= [-8 ± √80]/ 2(4)

= (-8 ± √(16 × 5))/8

x = (-8 + 4√5)/8, x = (-8 – 4√5)/8

x = 4(-2 + √5)/8, x = 4(-2 – √5)/8

x = (-2 + √5)/2, x = (-2 – √5)/2

Therefore, the real roots of the given quadratic equation are (-2 + √5)/2 and x = (-2 – √5)/2.

8. What is the nature of the roots of 5x 2 + 3x + 1 = 0? Find them.

5x 2 + 3x + 1 = 0

Here, a = 5, b = 3 and c = 1.

Let us calculate the discriminant to determine the nature of the roots of 5x 2 + 3x + 1 = 0.

= (3) 2 – 4(5)(1)

= 9 – 20

= -11 < 0

Thus, the given quadratic equation contains two imaginary roots.

= [-3 ± √(-11)]/ 2(5)

= (-3 ± i√11)/10

x = (-3 + i√11)/10, (-3 – i√11)/10

Hence, (-3 + i√11)/10 and (-3 – i√11)/10 are the roots of the given quadratic equation.

9. Solve: m 2 − 31 − 2m = −6 − 3m 2 − 4m

m 2 − 31 − 2m = −6 − 3m 2 − 4m

m 2 – 31 – 2m + 6 + 3m 2 + 4m = 0

4m 2 + 2m – 25 = 0

Comparing with the standard form of quadratic equation ax 2 + bx + c = 0, we get;

a = 4, b = 2 and c = -25

Now, b 2 – 4ac = (2) 2 – 4(4)(-25)

= 404 > 0

That means the given equation has two real and unequal roots.

m = [-2 ± √404]/ 2(4)

= [-2 ± √(4 × 101)]/8

= (-2 ± 2√101)/8

Therefore, m = (-2 + 2√101)/8 and m = (-2 – 2√101)/8.

10. Write the roots of the quadratic equation 3x 2 +3x = -3 using the quadratic formula.

3x 2 +3x = -3

3x 2 +3x + 3 = 0

3(x 2 + x + 1) = 0

x 2 + x + 1 = 0

Here, a = 1, b = 1 and c = 1.

= [-1 ± √{1 2 – 4(1)(1)}]/ 2(1)

= [-1 ± √(1 – 4)]/ 2

= (-1 ± √-3)/2

= (-1 ± i√3)/2

x = (-1 + i√3)/2, x = (-1 – i√3)/2

Therefore, the roots of 3x 2 + 3x = -3 are (-1 + i√3)/2 and (-1 – i√3)/2.

Practice Questions on Quadratic Formula

• Find the roots of the quadratic equation 6x 2 – 7x + 2 = 0.
• Find the roots of the quadratic equation x 2 – (√5/2)x – 1 = 0 using the quadratic formula.
• Determine the nature of the roots of 3 x 2 – 4√3 x + 4 = 0 quadratic equations. If the real roots exist, find them.
• What are the roots of the quadratic equation 6x 2 + 3 = -2x?
• Solve for x: 8x 2 + 7x − 8 = 0

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Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

• Factoring Quadratics
• Completing the Square
• Graphing Quadratic Equations
• The Quadratic Formula
• Online Quadratic Equation Solver

Each example follows three general stages:

• Take the real world description and make some equations
• Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

• t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

• h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

• $700,000 for manufacturing set-up costs, advertising, etc
• $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

• at $0, you just give away 70,000 bikes
• at $350, you won't sell any bikes at all
• at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

• Unit Sales = 70,000 − 200 x 230 = 24,000
• Sales in Dollars = $230 x 24,000 = $5,520,000
• Costs = 700,000 + $110 x 24,000 = $3,340,000
• Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

• Let x = the boat's speed in the water (km/h)
• Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

• when going upstream, v = x−2 (its speed is reduced by 2 km/h)
• when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation!

Let us solve it using our Quadratic Equation Solver .

• Enter 1, −1 and −6
• And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

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Quadratic equation is one of the most important and potentially high-scoring topics asked in different banking and insurance exams, such as IBPS PO, SBI PO, SBI Clerk, IBPS Clerk, RRB Assistant, RRB Scale 1, LIC Assistant, LIC AAO, etc. You can expect a set of 5 questions  on this topic in the prelims of every banking and insurance examination. Achieving proficiency in this topic demands keen observational skills. Nonetheless, with dedicated practice, you can attain mastery and achieve a perfect score in this area. Smartkeeda offers a diverse range of Quadratic Equation questions with solutions to facilitate effective practice and enhance your prospects of achieving a high score.  

Identifying Quadratic Patterns Question

A quadratic equation, denoted by the variable x, takes the form of ax 2  + bx + c = 0, where a, b, and c represent real numbers, with a ≠ 0. For instance, 2x 2  + x - 300 = 0 is a quadratic equation. However, to establish the standard representation of this equation, we arrange the terms of p(x) in descending order of their degrees, resulting in ax 2  + bx + c = 0, with a ≠ 0. This particular form, ax 2  + bx + c = 0, where 'a' is not equal to zero, is referred to as the standard form of a quadratic equation.  

Understanding the Quadratic Formula

Also known as the Sridharacharya formula , the quadratic formula is a formula that provides the two solutions to a quadratic equation. The Quadratic formula stands as the most straightforward method for determining the roots of a quadratic equation. In cases where certain quadratic equations resist easy factorization, the Quadratic formula offers a convenient and efficient means to swiftly calculate the roots.  

The Quadratic Formula

Solve using Quadratic formula 2x 2  - 7x + 3 = 0 Solution: Comparing the equation with the general form ax 2  + bx + c = 0 gives, a = 2, b = -7, and c = 3 Now, calculate the discriminant (b 2  - 4ac): b 2  - 4ac = (-7) 2  - 4 * 2 * 3 = 49 - 24 = 25 Now, substitute the values into the quadratic formula: x1 = (-b + √(b 2  - 4ac)) / (2a) x1 = (-(-7) + √25) / (2 * 2) x1 = (7 + 5) / 4 = 3 x2 = (-b - √(b 2  - 4ac)) / (2a) x2 = (-(-7) - √25) / (2 * 2) x2 = (7 - 5) / 4 = 1/2 So, the roots of the equation 2x 2 - 7x + 3 = 0 are x1 = 3 and x2 = 1/2.  

Factoring Quadratic Equations

Factoring quadratics is a technique used to represent the quadratic equation ax^2 + bx + c = 0 as a multiplication of its linear factors in the form (x - p) (x - q), where p and q represent the roots of the quadratic equation ax^2 + bx + c = 0. Factoring Techniques Quadratic equations can be factorized through various methods such as

• Splitting the middle term,
• Using quadratic formula or Shridharacharya formula
• Completing the square or square root method Understanding the pattern of Quadratic Equations asked in bank exams

Understanding the pattern of Quadratic Equations asked in bank exams

Bank exams often include these types of Quadratic Equation questions to assess candidates' problem-solving skills and their ability to discern relationships between variables, making it a critical component of the quantitative section. Questions on Quadratic Equations are asked in the form of inequalities in the Quantitative Aptitude section. Generally, two quadratic equations in two different variables are given. You have to solve both of the Quadratic equations to get to know the relation between both variables. Suppose we have two variables ‘x’ and ‘y’. The relationship between the variables can be any one of the following:   x > y x < y x = y or relation can’t be established between x & y x ≥ y x ≤ y

Practice Exercises

I.   x2 – 25x + 114 = 0 II.  y2 – 10y + 24 = 0

• if x > y
• if x < y
• if x = y or relationship between x and y can't be established

Tips for Efficient Quadratic Equation Solving

When solving quadratic equations, it's important to keep the following points in mind to ensure accurate and efficient problem-solving:

• Recognize that a quadratic equation is in the form ax^2 + bx + c = 0
• After finding potential solutions, ensure they satisfy the original equation.
• Carefully handle the signs (+/-) in the quadratic formula to avoid calculation errors.
• Observe carefully while comparing the roots of given equations

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Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

quadratic-equation-calculator

• High School Math Solutions – Quadratic Equations Calculator, Part 3 On the last post we covered completing the square (see link). It is pretty strait forward if you follow all the...

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Class 10 (Old)

Course: class 10 (old)   >   unit 4.

• Quadratic equations word problem: triangle dimensions
• Quadratic equations word problem: box dimensions
• Quadratic word problem: ball
• Word problems: Writing quadratic equations

Word problems: Solving quadratic equations

• Quadratic equations word problems (basic)
• Quadratic equations word problems (intermediate)
• Quadratic equations word problems (advanced)
• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍