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## Unit 14: Lesson 6

- The quadratic formula
- Understanding the quadratic formula
- Worked example: quadratic formula (example 2)
- Worked example: quadratic formula (negative coefficients)
- Using the quadratic formula: number of solutions
- Quadratic formula review
- Discriminant review

## Quadratic formula

- (Choice A) x = 3 , − 1 2 x =3,-\frac{1}{2} x = 3 , − 2 1 x, equals, 3, comma, minus, start fraction, 1, divided by, 2, end fraction A x = 3 , − 1 2 x =3,-\frac{1}{2} x = 3 , − 2 1 x, equals, 3, comma, minus, start fraction, 1, divided by, 2, end fraction
- (Choice B) x = 5 ± 57 16 x =\dfrac{5 \pm \sqrt{57}}{16} x = 1 6 5 ± 5 7 x, equals, start fraction, 5, plus minus, square root of, 57, end square root, divided by, 16, end fraction B x = 5 ± 57 16 x =\dfrac{5 \pm \sqrt{57}}{16} x = 1 6 5 ± 5 7 x, equals, start fraction, 5, plus minus, square root of, 57, end square root, divided by, 16, end fraction
- (Choice C) x = 1 ± 17 − 4 x =\dfrac{1 \pm \sqrt{17}}{-4} x = − 4 1 ± 1 7 x, equals, start fraction, 1, plus minus, square root of, 17, end square root, divided by, minus, 4, end fraction C x = 1 ± 17 − 4 x =\dfrac{1 \pm \sqrt{17}}{-4} x = − 4 1 ± 1 7 x, equals, start fraction, 1, plus minus, square root of, 17, end square root, divided by, minus, 4, end fraction
- (Choice D) x = − 4 ± 34 3 x =\dfrac{-4 \pm \sqrt{34}}{3} x = 3 − 4 ± 3 4 x, equals, start fraction, minus, 4, plus minus, square root of, 34, end square root, divided by, 3, end fraction D x = − 4 ± 34 3 x =\dfrac{-4 \pm \sqrt{34}}{3} x = 3 − 4 ± 3 4 x, equals, start fraction, minus, 4, plus minus, square root of, 34, end square root, divided by, 3, end fraction

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About the quadratic formula, quadratic formula video lesson.

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## Section 2.5 : Quadratic Equations - Part I

For problems 1 – 7 solve the quadratic equation by factoring.

- \({u^2} - 5u - 14 = 0\) Solution
- \({x^2} + 15x = - 50\) Solution
- \({y^2} = 11y - 28\) Solution
- \(19x = 7 - 6{x^2}\) Solution
- \(6{w^2} - w = 5\) Solution
- \({z^2} - 16z + 61 = 2z - 20\) Solution
- \(12{x^2} = 25x\) Solution

For problems 8 & 9 use factoring to solve the equation.

For problems 10 – 12 use factoring to solve the equation.

- \(\displaystyle \frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\) Solution
- \(\displaystyle \frac{{4z}}{{z + 1}} + \frac{5}{z} = \frac{{6z + 5}}{{{z^2} + z}}\) Solution
- \(\displaystyle x + 1 = \frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}\) Solution

For problems 13 – 16 use the Square Root Property to solve the equation.

- \(9{u^2} - 16 = 0\) Solution
- \({x^2} + 15 = 0\) Solution
- \({\left( {z - 2} \right)^2} - 36 = 0\) Solution
- \({\left( {6t + 1} \right)^2} + 3 = 0\) Solution

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## How to Solve Quadratic Equations

Last Updated: February 10, 2023 References Approved

## Factoring the Equation

- 3x = -1 ..... by subtracting
- 3x/3 = -1/3 ..... by dividing
- x = -1/3 ..... simplified
- x = 4 ..... by subtracting
- x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

## Using the Quadratic Formula

## Completing the Square

## Practice Problems and Answers

## Expert Q&A

- If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source ⧼thumbs_response⧽ Helpful 0 Not Helpful 0
- As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. ⧼thumbs_response⧽ Helpful 0 Not Helpful 0
- If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. ⧼thumbs_response⧽ Helpful 0 Not Helpful 0

## You Might Also Like

- ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
- ↑ http://www.mathsisfun.com/algebra/factoring-quadratics.html
- ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
- ↑ https://www.cuemath.com/algebra/quadratic-equations/
- ↑ https://www.purplemath.com/modules/solvquad4.htm
- ↑ http://www.purplemath.com/modules/quadform.htm
- ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
- ↑ http://www.mathsisfun.com/algebra/completing-square.html
- ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

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## Solving quadratic equations

## Quadratic equations

Here are some examples of quadratic equations in this form:

- \(2x^2 - 2x - 3 = 0\) . \(a = 2\) , \(b = -2\) and \(c = -3\)
- \(2x(x + 3) = 0\) . \(a = 2\) , \(b = 6\) and \(c = 0\) (in this example, the bracket can be expanded to \(2x^2 + 6x = 0\) )
- \((2x + 1)(x - 5) = 0\) . \(a = 2\) , \(b = -9\) and \(c = -5\) (this will expand to \(2x^2 - 9x - 5 = 0\) )
- \(x^2 + 2 = 4\) . \(a = 1\) , \(b = 0\) and \(c = -2\)
- \(3x^2 = 48\) . \(a = 3\) , \(b = 0\) and \(c = -48\) (in this example \(c = -48\) , but has been rearranged to the other side of the equation)

\(3x^2 = 48\) is an example of a quadratic equation that can be solved simply.

Square root both sides to isolate x

## Solving by factorising

The product of \(x\) and \(x + 3\) is 0, so \(x = 0\) or \(x + 3 = 0\) , or both.

\[\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}\]

Solve \((x + 1)(x - 5) = 0\) .

The product of \(x + 1\) and \(x - 5\) is 0, so one or both brackets must also be equal to 0.

\[\begin{array}{rcl} x + 1 & = & 0 \\ -1 && -1 \\ x & = & -1 \end{array}\]

\[\begin{array}{rcl} x - 5 & = & 0 \\ + 5 && + 5 \\ x & = & 5 \end{array}\]

Begin by factorising the quadratic .

The quadratic will be in the form \((x + a)(x + b) = 0\) .

Find two numbers with a product of 12 and a sum of 7.

\(3 \times 4 = 12\) , and \(3 + 4 = 7\) , so \(a\) and \(b\) are equal to 3 and 4. This gives:

The product of \(x + 3\) and \(x + 4\) is 0, so \(x + 3 = 0\) or \(x + 4 = 0\) , or both.

\[\begin{array}{rcl} x + 3 & = & 0 \\ - 3 && - 3 \\ x & = & -3 \end{array}\]

\[\begin{array}{rcl} x + 4 & = & 0 \\ - 4 && - 4 \\ x & = & -4 \end{array}\]

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## COMMENTS

If you have a general quadratic equation like this: ax^2+bx+c=0 ax2 + bx + c = 0 Then the formula will help you find the roots of a quadratic equation, i.e. the values of x x where this equation is solved. The quadratic formula x=\dfrac {-b\pm\sqrt {b^2-4ac}} {2a} x = 2a−b ± b2 − 4ac It may look a little scary, but you'll get used to it quickly!

The quadratic formula Quadratic formula CCSS.Math: HSA.REI.B.4, HSA.REI.B.4b Google Classroom Solve. 6 + 2x^2 - 3x = 8x^2 6 + 2x2 −3x = 8x2 Choose 1 answer: x =3,-\frac {1} {2} x = 3,−21 A x =3,-\frac {1} {2} x = 3,−21 x =\dfrac {5 \pm \sqrt {57}} {16} x = 165 ± 57 B x =\dfrac {5 \pm \sqrt {57}} {16} x = 165 ± 57

How to Solve a Quadratic Equation Using the Quadratic Formula Example 9.3. 1 Solve 2 x 2 + 9 x − 5 = 0 by using the Quadratic Formula. Answer Example 9.3. 2 Solve 3 y 2 − 5 y + 2 = 0 by using the Quadratic Formula. Answer Example 9.3. 3 Solve 4 z 2 + 2 z − 6 = 0 by using the Quadratic Formula. Answer

Remember, the quadratic formula looks like this: ± √ ( Our coefficients: , , and Our equation after inserting the coefficients: ± √ ( 5 Use the order of operations to simplify the formula. Download Article Just do the math in the equation as you normally would.

Quadratic Formula Calculator Watch on Example (Click to try) 2 x 2 − 5 x − 3 = 0 About the quadratic formula Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: x = Quadratic Formula Video Lesson Solve with the Quadratic Formula Step-by-Step [1:29] Need more problem types? Try MathPapa Algebra Calculator Show Keypad

This video explains how to solve quadratic equations using the quadratic formula. My Website: https://www.video-tutor.net How To Use The Quadratic Formula To Solve Equations How to...

For problems 1 - 7 solve the quadratic equation by factoring. u2 −5u−14 = 0 u 2 − 5 u − 14 = 0 Solution x2 +15x =−50 x 2 + 15 x = − 50 Solution y2 = 11y−28 y 2 = 11 y − 28 Solution 19x = 7−6x2 19 x = 7 − 6 x 2 Solution 6w2 −w =5 6 w 2 − w = 5 Solution z2 −16z +61 = 2z −20 z 2 − 16 z + 61 = 2 z − 20 Solution 12x2 = 25x 12 x 2 = 25 x Solution

Solve by using the Quadratic Formula: x(x + 6) + 4 = 0. Solution: Our first step is to get the equation in standard form. Distribute to get the equation in standard form. This equation is now in standard form. Identify the values of a, b, c. Write the Quadratic Formula. Then substitute in the values of a, b, c. Simplify. Simplify the radical.

There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps. Method 1 Factoring the Equation 1

Learn about quadratic equations using our free math solver with step-by-step solutions. Skip to main content. Microsoft Math Solver. Solve Practice Download. ... Type a math problem. Type a math problem. Solve. Examples. x^2-3x=28. x ^ { 2 } - 5 x + 3 y = 20. x^2-10x+25=0. 2x^2+12x+40=0 ...

In algebra, a quadratic equation (from Latin quadratus 'square') is any equation that can be rearranged in standard form as where x represents an unknown value, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.) The numbers a, b, and c are the coefficients of the equation ...

Use the quadratic formula to find the solutions. −b±√b2 −4(ac) 2a - b ± b 2 - 4 ( a c) 2 a Substitute the values a = 1 a = 1, b = 7 b = 7, and c = 10 c = 10 into the quadratic formula and solve for x x. −7±√72 −4 ⋅(1⋅10) 2⋅1 - 7 ± 7 2 - 4 ⋅ ( 1 ⋅ 10) 2 ⋅ 1 Simplify. Tap for more steps... x = −7±3 2 x = - 7 ± 3 2

This algebra video tutorial explains how to solve quadratic equations by factoring in addition to using the quadratic formula. This video contains plenty of examples and practice...

There are many ways to solve quadratics. All quadratic equations can be written in the form \ (ax^2 + bx + c = 0\) where \ (a\), \ (b\) and \ (c\) are numbers (\ (a\) cannot be equal to 0, but...

A demonstration of the techniques for finding the zeros or roots of a quadratic function to solve a quadratic equation on the Ti84 graphing Homework Support Solutions Solving math equations can be challenging, but it's also a great way to improve your problem-solving skills.