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9.6 Solving Quadratic Equations by Using Square Roots

Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

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Page 1: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

9.6Solving Quadratic

Equations by Using Square Roots

Page 2: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Warm UpFind each square root.

Solve each equation.

5. –6x = –60 6.

7. 2x – 40 = 0 8. 5x = 3

6 11

–25

1. 2.

3. 4.

x = 10 x = 80

x = 20

Page 3: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

California Standards

2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules for exponents.23.0 Students apply

quadratic equations to physical problems, such as the motion of an object under the force of gravity.

Page 4: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Some quadratic equations cannot be easily solved by factoring. Square roots can be used to solve some of these quadratic equations. Recall from Lesson 1-5 that every positive real number has two square roots, one positive and one negative. (Remember also that the symbol indicates a nonnegative square root.)

Page 5: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Negative square root of 9

Positive and negativesquare roots of 9

Positive square root of 9

When you take the square root to solve an equation, you must find both the positive and negative square root. This is indicated by the symbol ±√ .

REVIEW

Page 6: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

The expression ±3 means “3 or –3” and is read “plus or minus three.”

Reading Math

Page 7: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20
Page 8: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Solve using square roots.

x2 = 169

x = ± 13

The solutions are 13 and –13.

Solve for x by taking the square root of both sides. Use ± to show both square roots.

Substitute 13 into the original equation.Check x2 = 169

(13)2 169 169 169

Page 9: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Solve using square roots.

x2 = –49 There is no real number whose square is negative.

There is no real solution. The solution set is the empty set, ø.

Page 10: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Now you trySolve using square roots

1. x2 = 121

2. x2 = -25

3. x2 = 0

4. x2 = -16

5. x2 = 100

6. x2 = 64

x = ±11no real solution. ø

x = 0

no real solution. ø

x = ±10

x = ±8

Page 11: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

If a quadratic equation is not written in the form x2 = a, use inverse operations to isolate x2 before taking the square root of both sides.

Page 12: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Solve using square roots.

x2 + 7 = 32

–7 –7 x2 + 7 = 32

x2 = 25

Subtract 7 from both sides.

Take the square root of both sides.

Page 13: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

16x2 – 49 = 0

16x2 – 49 = 0

+49 +49

Add 49 to both sides.

Divide by 16 on both sides.

Take the square root of both sides. Use ± to show both square roots.

Solve by using square roots.

Page 14: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

36x2 = 1Divide by 36 on both sides.

Take the square root of both sides. Use ± to show both square roots.

Solve by using square roots.

Page 15: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Solve by using square roots.

100x2 + 49 = 0

100x2 + 49 = 0–49 –49

100x2 =–49

There is no real number whose square is negative.

Subtract 49 from both sides.

Divide by 100 on both sides.

ø

Page 16: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Solve. Round to the nearest hundredth.

0 = 90 – x2

+ x2 + x2 0 = 90 – x2

x2 = 90

Add x2 to both sides.

Take the square root of both sides.

Estimate

The exact solutions are and The approximate solutions are 9.49 and –9.49.

.

Page 17: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Now you trySolve using square roots

1. x2 -196 = 0

2. 0 = 3x2 -48

3. 24x2 +96 = 0

4. 10x2 - 75= 15

5. 0 = 4x2 +144

6. 5x2 – 105 = 20

x = ±14

no real solution. ø

x = ±3

x = ±5

x = ±4

no real solution. ø

Page 18: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Application Ms. Pirzada is building a retaining wall along one of the long sides of her rectangular garden. The garden is twice as long as it is wide. It also has an area of 578 square feet. What will be the length of the retaining wall?

Let x represent the width of the garden.

lw = AUse the formula for area of a rectangle.

Substitute x for w, 2x for l, and 578 for A.

2x x = 578 ●

l = 2w

2x2 = 578

Length is twice the width.

Page 19: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Continued

2x2 = 578

x = ±17

Take the square root of both sides.

Negative numbers are not reasonable for width, so x = 17 is the only solution that makes sense. Therefore, the length is 2w or 34 feet.

Divide both sides by 2.

Page 20: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Lesson Quiz: Part I

Solve using square roots. Check your answers.

1. x2 – 195 = 1

2. 4x2 – 18 = –9

3. 2x2 – 10 = –12

4. Solve 0 = –5x2 + 225. Round to the nearest hundredth.

± 14

± 6.71

ø

Page 21: Warm Up Find each square root. Solve each equation. 5. –6x = –606. 7. 2x – 40 = 08. 5x = 3 611 –25 1. 2. 3. 4. x = 10 x = 80 x = 20

Lesson Quiz: Part II

5. A community swimming pool is in the shape of a trapezoid. The height of the trapezoid is twice as long as the shorter base and the longer base is twice as long as the height.

The area of the pool is 3675 square feet. What is the length of the longer base? Round to the nearest foot.

(Hint: Use )

108 feet