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**If b2 = a, then b is a square root of a.**

9.1 Square Roots SQUARE ROOT OF A NUMBER If b2 = a, then b is a square root of a. Examples: 32 = 9, so 3 is a square root of 9. (-3)2 = 9, so -3 is a square root of 9.

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**Evaluate the expression. - 𝟒**

Chapter 9 Test Review Evaluate the expression. - 𝟒

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**Evaluate the expression. 𝟏𝟒𝟒**

Chapter 9 Test Review Evaluate the expression. 𝟏𝟒𝟒

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**Evaluate the expression. 𝟏𝟎𝟎**

Chapter 9 Test Review Evaluate the expression. 𝟏𝟎𝟎

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**Evaluate the expression. - 𝟐𝟓**

Chapter 9 Test Review Evaluate the expression. - 𝟐𝟓

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**9.2 Solving Quadratic Equations by Finding Square Roots**

When b = 0, this equation becomes ax2 + c = 0. One way to solve a quadratic equation of the form ax2 + c = 0 is to isolate the x2 on one side of the equation. Then find the square root(s) of each side.

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Chapter 9 Test Review Solve the equation. x2 = 144

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Chapter 9 Test Review Solve the equation. 8x2 = 968

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Chapter 9 Test Review Solve the equation. 5x2 – 80 = 0

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Chapter 9 Test Review Solve the equation. 3x2 – 4 = 8

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**9.3 Simplifying Radicals PRODUCT PROPERTY OF RADICALS ab = a ∙ b**

EXAMPLE: = 4∙5 = ∙ 5 = 2 5

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**Simplify the expression. 𝟒𝟓**

Chapter 9 Test Review Simplify the expression. 𝟒𝟓

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**Simplify the expression. 𝟐𝟖**

Chapter 9 Test Review Simplify the expression. 𝟐𝟖

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**Simplify the expression. 36 24**

Chapter 9 Test Review Simplify the expression. 36 24

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**Simplify the expression. 8 6**

Chapter 9 Test Review Simplify the expression. 8 6

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**9.5 Solving Quadratic Equations by Graphing**

The x-intercepts of graph y = ax2 + bx + c are the solutions of the related equations ax2 + bx + c = 0. Recall that an x-intercept is the x-coordinate of a point where a graph crosses the x-axis. At this point, y = 0.

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Chapter 9 Test Review Use a graph to estimate the solutions of the equation. Check your solutions algebraically. x2 – 3x = -2

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Chapter 9 Test Review Use a graph to estimate the solutions of the equation. Check your solutions algebraically. -x2 + 6x = 5

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Chapter 9 Test Review Use a graph to estimate the solutions of the equation. Check your solutions algebraically. x2 – 2x = 3

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**9.6 Solving Quadratic Equations by the Quadratic Formula**

The solutions of the quadratic equation ax2 + bx + c = 0 are: x = −𝑏 ± 𝑏 2 −4𝑎𝑐 2𝑎 when a ≠ 0 and b2 – 4ac > 0.

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**Use the quadratic formula to solve the equation. 3x2 – 4x + 1 = 0**

Chapter 9 Test Review Use the quadratic formula to solve the equation. 3x2 – 4x + 1 = 0

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**Use the quadratic formula to solve the equation. -2x2 + x + 6 = 0**

Chapter 9 Test Review Use the quadratic formula to solve the equation. -2x2 + x + 6 = 0

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**Use the quadratic formula to solve the equation. 10x2 – 11x + 3 = 0**

Chapter 9 Test Review Use the quadratic formula to solve the equation. 10x2 – 11x + 3 = 0

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**9.7 Using the Discriminant**

In the quadratic formula, the expression inside the radical is the DISCRIMINANT. x = −𝑏 ± 𝑏 2 −4𝑎𝑐 2𝑎 DISCRIMINANT 𝐛 𝟐 - 4ac

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Chapter 9 Test Review Find the value of the discriminant. Then determine whether the equation has two solutions, one solution, or no real solution. 3x2 – 12x + 12 =0

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Chapter 9 Test Review Find the value of the discriminant. Then determine whether the equation has two solutions, one solution, or no real solution. 2x2 + 10x + 6 =0

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Chapter 9 Test Review Find the value of the discriminant. Then determine whether the equation has two solutions, one solution, or no real solution. -x2 + 3x - 5 =0

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