Integrated Math II Name: ________________________________________ Period: ______ Date: _______________

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5.2.1 Perfect Square Equations

Simplifying Square Roots

A square root is simplified when there are no more perfect square factors (square numbers such as 4, 25, and 81) under the radical sign. Here are some examples of this method.

Example 1: Simplify 45

First rewrite 45 in an equivalent factored form so that one of the factors is a perfect square. Simplify the square root of the perfect square. Verify with your calculator that both 3 5 and 45 ≈ 6.71.

Example 2: Note that 72 can be rewritten as 36 ∙ 2, or 8 ∙9 and you still get the same result when you simplify completely.

NOTE: 45 is in exact form, 3 5 is in simplified radical form, and 6.71 is a decimal approximation of these numbers.

In Section 5.1, you made connections between graphs, tables, situations, and equations for quadratic functions. To locate the x-intercepts of a parabola y = ax2 + bx + c, you solved the related quadratic equation, 0 = ax2 + bx + c, by factoring and using the Zero Product Property. In this section, you will focus on solving quadratic equations. Sometimes the form of a quadratic equation can give you a fast way to figure out the solutions! 5-60. Without rewriting, determine the value(s) of x that make(s) each equation true.

a. (x – 1)2 = 4 (𝒙 − 𝟏)𝟐 = 𝟒

𝒙 − 𝟏 = −𝟐 or 𝒙 − 𝟏 = 𝟐 𝒙 = −𝟏 or 𝒙 = 𝟑 Two solutions

b. (x – 1)2 = 0 (𝑥 − 1)! = 0 𝑥 − 1 = 0 𝑥 = 1

c. (x – 1)2 = –4 (𝒙 − 𝟏)𝟐 = −𝟒

No solution because −𝟒 is an imaginary number

5-61. The quadratic equation (x – 3)2 = 12 is written in perfect square form. It is called this because the expression (x – 3)2 forms a square when built with algebra tiles.

Integrated Math II Name: ________________________________________ Period: ______ Date: _______________

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a. Solve (x – 3)2 = 12. Write your answer in exact form (or radical form). That is, write it in a form that is precise and does not have any rounded decimals. 𝑥 − 3 ! = 12

𝑥 − 3 = 2 3 or 𝑥 − 3 = −2 3 𝑥 = 3 + 12 or 𝑥 = 3 − 2 3

b. How many solutions are there? Justify your answer.

There are two solutions because the square root of 12 has two solutions.

c. The solution(s) from part (a) are irrational. That is, they are decimals that never repeat and never end. Write the solution(s) in approximate decimal form. Round your answers to the nearest hundredth.

𝑥 = 3 + 12 ≈ 6.46 or 𝑥 = 3 − 2 3 ≈–0.46

5-62. THE NUMBER OF SOLUTIONS. The equation in problem 5-61 had two solutions. However, in problem 5-60, you saw that a quadratic equation might have one solution or no solutions at all. How can you determine the number of solutions to a quadratic equation? With your team, solve the equations below. Express your answers in both exact form and approximate form. Look for patterns among the equations with no solution and those with only one solution.

a. (5 – 10x)2 = 0 (5 − 10𝑥)! = 0

5 − 10𝑥 = 0

𝑥 =12

One solution

b. (x + 2)2 = –10 (𝑥 + 2)! = −10

No solution because −10 is an imaginary number

c. (2x – 3)2 = 49 (2𝑥 − 3)! = 49

2𝑥 − 3 = 7 or 2𝑥 − 3 = −7 𝑥 = 5 or 𝑥 = −2 Two solutions

d. (7x – 5)2 = –2 (𝟕𝒙 − 𝟓)𝟐 = −𝟐

No solution because −𝟐 is an imaginary number

e. (x + 11)2 + 5 = 5 (𝒙 + 𝟏𝟏)𝟐 = 𝟎 (𝒙 + 𝟏𝟏)𝟐 = 𝟎 𝒙 + 𝟏𝟏 = 𝟎 𝒙 = −𝟏𝟏

One solution

f. (x + 4)2 = 20 (𝒙 + 𝟒)𝟐 = 𝟐𝟎

𝒙 + 𝟒 = 𝟐 𝟓 or 𝒙 + 𝟒 = −𝟐 𝟓 𝒙 = −𝟒 + 𝟐 𝟓 ≈ 𝟎.𝟒𝟕 or 𝒙 = −𝟒 − 𝟐 𝟓 ≈ −𝟖.𝟑𝟕 Two solutions

5-63. Use the patterns you found in problem 5-62 to determine how many solutions each quadratic below has. You do not need to solve the equations.

a. (5x – 2)2 + 6 = 0 (5𝑥 − 2)! = −6

No solution because −6 is an imaginary number

b. (4 + 2x)2 = 0 (4 + 2𝑥)! = 0

One solution

c. 11 = (7 + 2x)2 (7 + 2𝑥)! = 11

7 + 2𝑥 = 11 or 7 + 4𝑥 =− 11 Two solutions

, 0) so y = a (x + ½)(x – 3/2) = a (x2 – x –¾) = 4 (x2 – x – ¾) = 4x2 – 4x – 3.