Ch 9: Quadratic EquationsB) Square Roots

Objective:

To solve quadratic equations using square roots.

Quadratic Expression

An expression in which 2 is the largest exponent.

ax2 + bx + c

Quadratic Equation

An equation in which 2 is the largest exponent.

ax2 + bx + c = 0

Square root of a quadratic

The square root of a variable squared (x2) equals the absolute value of the square root.

Definitions

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x 2 = |x| = ± x

1) Isolate x2 (get x2 on one side and the number on the other side)

2) Take the square root of BOTH sides (keep the equation balanced)

3) Solve for the absolute value of x (this creates 2 equations)

Look to see if the number is on the diagonal of the multiplication table.

(a) If so, it is a perfect square and you have your answer. (Don’t forget the ± symbol)

1) If not, simplify the radicand and solve for both equations

Note: There should be two answers!

Rules

Multiplication Tablex 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

Perfect Squares

14

916

2536

4964

81100

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x2 = 9

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x2 = 9

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x = 3

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x = ±3

Two Solutions

x2 = 12

√x2 = √ 2 2 3|x| = 2√3

x = ±2√3

Two Solutions

x2 = -9

√x2 = √-9

No Real solution

√x2 = √12

2 6

2 3

On the Diagonal Not on the Diagonal Negative on the inside

Example 1 Example 2 Example 3

y2 + 5 = 10- 5 - 5y2 = 5

Example 4 Example 5 Example 6

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y 2 = 5

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| y |= 5

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y = ± 5

2m2 − 3 = 5+3 +3

2m2 = 8

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m2 = 4

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| m |= 2

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m = ±2

2 2m2 = 4

3r2 + 7 = 8−7 −7

3r2 = 13 3

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r2 =1

3

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r2 =1

3

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| r |=1

3

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r = ±1

3

y2 + 4 = 2

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y +1( )2

= 16

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x − 3( )2

= 25

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y +1 = ±4

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y +1( )2

= 16

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y +1 = 4

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−1 −1

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y = −1 ± 4

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−1+ 4

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−1− 4

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3

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or − 5

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x − 3 = ±5

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x − 3( )2

= 25

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x − 3 = 5

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+3 + 3

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3 + 5

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3− 5

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8

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or − 2

x = 3 ± 5

- 4 - 4

y2 = -2

No Real Solution

Example 7 Example 8 Example 9

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y 2 = −2

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x2 = 36 x2 = 18 x2 – 4 = 12

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x2 = 36

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x = 6

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x = ±6

√x2 = √18

|x| = √233

x = ± 3√2

+ 4 + 4

x2 = 16

√x2 = √16

|x| = ± 4

1) 2) 3)

Classwork

4) 5)

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y2 = 7

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m2 − 6 = 7

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y2 = 7

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y = 7

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y = ± 7

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+6 + 6

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m2 =13

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m2 = 13

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m = 13

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m = ± 13

6)

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2x2 −16 = 4

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+16 +16

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2x2 = 20

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x2 = 10

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x2 = 10

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x = 10

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x = ± 10

2 2

7)

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3y2 − 20 = 28

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y2 = 16

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y = 4

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y = ±4

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+20 + 20

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3y2 = 48

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3 3

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y2 = 16

8)

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2y2 + 30 =16

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−30 − 30

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2y2 = −14

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2 2

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y2 = −7

No Real Solution€

y 2 = −7