Math 106 Worksheets: Quadratic Equations. Quadratic Equations: Solving Quadratic Equations with Square Roots 1. Solving Quadratic Equations with Square Roots 2. Factoring Quadratic Expressions. Solving Quadratic Equations By Factoring. Completing The Square 1. Completing The Square 2. Solving Equations With Completing The Square 1Math 154b Solving Using The Quadratic Formula Worksheet Answers. July 19, 2020 via admin. 21 Posts Related to Math 154b Solving Using The Quadratic Formula Worksheet Answers. What Is A Metaphor Math Worksheet Quadratic Formula Answers. Quadratic Formula Word Problems Worksheet Answers.To remedy quadratic equations that are unfactorable, you can need to understand how to make use of the quadratic formula, and this quiz/worksheet combo will permit you to better perceive this formula.Quadratic Formula Kuta - Displaying most sensible Eight worksheets found for this idea.. Some of the worksheets for this concept are Solve every equation with the quadratic, Solve each and every equation with the quadratic, Using the quadratic formula kuta tool algebra 2, Quadratic formula discriminant date period, Quadratic equations by way of finishing the sq., Understanding the discriminant date lengthUse the Quadratic Formula Name:_____ to Solve the Equations http://math.about.com Score: /10

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We are suggesting you to refer the Quadratic Formula Worksheet with Answers and start preparation. Mostly use the factorization technique to get the roots of the equation simply.

1. Which of the following are quadratic equations?

(i) 5x² + √8x + 5 = 0

(ii) x + 6/x = x

(iii) x² = 0

(iv) x² – 4x = 0

(v) x² – 25 = 0

Solution:(i) 5x² + √8x + 5 = 0 is in the type of ax² + bx + c = 0. So, it is a quadratic equation.

(ii) x + 6/x = x

(x² + 6)/x = x

x² + 6 = x²

Which is not in the form of ax² + bx + c = 0. So, it's not a quadratic equation.

(iii) x² = Zero is in the form of ax² + bx + c = 0. So, it is a quadratic equation.

(iv) x² – 4x = 0 is in the form of ax² + bx + c = 0. So, this can be a quadratic equation.

(v) x² – 25 = Zero is in the type of ax² + bx + c = 0. So, it is a quadratic equation.

2. Find which of the following are quadratic equations?

(i) x² = 64

(ii) x² + 4x + 2 = 0

(iii) x² + 3/x² = 8

(iv) x (x + 1) – (x + 2) (x + 2) = -8

(v) x² – √x + 4 = 0

Solution:(i) x² = Sixty four will also be written as x² – 64 = 0

Which is in the type of ax² + bx + c = 0. So, it is a quadratic equation.

(ii) x² + 4x + 2 = Zero is in the type of ax² + bx + c = 0. So, this is a quadratic equation.

(iii) x² + 3/x² = 8

(x⁴ + 3) /x² = 8

x⁴ + 3 = 8x²

Which is not in the type of ax² + bx + c = 0. So, it isn't a quadratic equation.

(iv) x (x + 1) – (x + 2) (x + 2) = -8

x² + x – [x² + 4x + 4] = -8

x² + x – x² – 4x – 4 = -8

-3x – 4 = -8

3x = -4 + 8 = 4

Which is not in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

(v) x² – √x + 4 = Zero isn't in the form of ax² + bx + c = 0. So, it is not a quadratic equation.

3. Find if the given values are the answer of the following equations:

(i) x² – 6√5x + 25 = 0; x = 5 and x = 5√5

(ii) 8x² + 2x – 1 = 0; x = 1/2 and x = -1/4

(iii) x² – 3x = 40; x = -5, x = 8

(iv) 6x² – 13x + 5 = 0; x = 1/2, and x = 5/3

Solution:(i) Given equation is x² – 6√5x + 25 = 0

a = 1, b = -6√5, c = 25

roots (α, β) = [-b ± √(b² – 4ac)] / 2a

= [-(-6√5) ± √((-6√5)² – 4 * 1 * 25)] / (2 * 1)

= [6√5 ± √(180 – 100)] / 2

= [6√5 ± √80] / 2

= [6√5 ± 4√5] / 2

= [6√5 + 4√5] / 2 and [6√5 – 4√5] / 2

= 10√5 / 2 and a couple of√5 / 2

= 5√5 and √5

So, the roots are pleasurable the equation.

(ii) Given quadratic equation is 8x² + 2x – 1 = 0

8x² + 4x – 2x – 1 = 0

= 4x(2x + 1) -1(2x + 1) = 0

(2x + 1)(4x – 1) = 0

2x + 1 = 0 and 4x – 1 = 0

2x = -1 and 4x = 1

x = -1/2 and x = 1/4

Therefore, the given roots are correct.

(iii) Given quadratic equation is x² – 3x = 40

It can be written as x² – 3x – 40 = 0

By using the factorization method

x² – 8x + 5x – 40 = 0

x(x – 8) + 5(x – 8) = 0

(x + 5)(x – 8) = 0

x – 8 = Zero and x + 5 = 0

x = 8 and x = -5

Therefore, the given roots are right kind.

(iv) Given main points are 6x² – 13x + 5 = 0; x = 1/2, and x = 5/3

6x² – 10x – 3x + 5 = 0

2x(3x – 5) – 1(3x – 5) = 0

(2x – 1)(3x – 5) = 0

2x – 1 = 0 and 3x – 5 = 0

2x = 1 and 3x = 5

x = 1/2 and x = 5/3

Therefore, the given roots are correct.

4. Solve the following quadratic equations and to find the solution.

(i) x² – (√2 – √3)x – √6 = 0

(ii) 5x² + 49x + 72 = 0

(iii) x² + 2x – 24 = 0

Solution:(i) Given quadratic equation is x² – (√2 – √3)x – √6

a = 1, b = √3 – √2, c = -√6

Quadratic equation roots formula is given as

roots (α, β) = [-b ± √(b² – 4ac)] / 2a

= [-(√3 – √2) ± √((√3 – √2)² – 4 * 1 * (-√6)] / 2 * 1

= [(√2 – √3) ± √(5 – 2√6 + 4√6)] / 2

= [(√2 – √3) ± √(5 + 2√6)] / 2

= [(√2 – √3) + √(5 + 2√6)] / 2 and [(√2 – √3) – √(5 + 2√6)] / 2

= [1.141 – 1.732 + √(5 + 2(2.44)] / 2 and [1.141 – 1.732 – √(5 + 2(2.44)] / 2

= [-0.591 + √(5 + 4.88)] / 2 and [-0.591 – √(5 + 4.88)] / 2

= [-0.591 + √9.88] / 2 and [-0.591 – √9.88] / 2

= [-0.591 + 3.143] / 2 and [-0.591 – 3.143] / 2

= 2.55 / 2 and -3.734 / 2

= 1.275 and -1.875

= √2 and -√3

(ii) Given equation is 5x² + 49x + 72 = 0

5x² + 40x + 9x + 72 = 0

5x(x + 8) +9(x + 8) = 0

(5x + 9) (x + 8) = 0

(5x + 9) = 0 and (x + 8) = 0

5x = -9 and x = -8

x = -9/5

The roots are (-8, -9/5)

(iii) Given equation is x² + 2x – 24 = 0

x² + 6x – 4x – 24 = 0

x(x + 6) -4(x + 6) = 0

(x – 4)(x + 6) = 0

(x – 4) = Zero and (x + 6) = 0

x = Four and x = -6

The roots are (4, -6).

5. Find the roots for the following quadratic equations:

(i) 50x² + 110x + 48 = 0

(ii) 4x² + 12x + 9 = 0

(iii) 16x² – 49 = 0

Solution:(i) Given quadratic equation is 50x² + 110x + 48 = 0

50x² + 80x + 30x + 48 = 0

10x(5x + 8) + 6(5x + 8) = 0

(10x + 6) (5x + 8) = 0

(10x + 6) = 0 and (5x + 8) = 0

10x = -6 and 5x = -8

x = -6/10 and x = -8/5

Roots are (-3/5, -8/5)

(ii) Given quadratic equation is 4x² + 12x + 9 = 0

4x² + 6x + 6x + 9 = 0

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

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

(2x + 3) = Zero and (2x + 3) = 0

2x = -3 and 2x = -3

x = -3/2 and x = -3/2

The roots are (-3/2, -3/2)

(iii) Given quadratic equation is 16x² – 49 = 0

(4x)² – 7² = 0

(4x)² = 7²

Applying square root on all sides

√(4x)² = √7²

4x = ±7

x = 7/ 4 and x = -7/4

The roots are (-7/4, 7/4).

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