1. Solving Quadratic Equations by Using Square Roots 8-7
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 1
Holt Algebra 1

2. Warm Up
Find each square root.
Solve each equation.
5. –6x = –60 6.
7. 2x – 40 = x = 3 1. 6 2. 11 3.
–25 4.
x = 10
x = 80
x = 20

3. Objective
Solve quadratic equations by using square roots.

4. Some quadratic equations cannot be easily solved by factoring
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.

5. Positive
Square root of 9
Negative
Square root of 9
When you take the square root of a positive number and the sign of the square root is not indicated, you must find both the positive and negative square root. This is indicated by ±√
Positive and negative
Square roots of 9

6. The expression ±3 is read “plus or minus three”
Reading Math

8.   Example 1A: Using Square Roots to Solve x2 = a
Solve using square roots. Check your answer.
x2 = 169
Solve for x by taking the square root of both sides. Use ± to show both square roots.
x = ± 13
The solutions are 13 and –13.
Check x2 = 169
(13) 
x2 = 169
(–13) 
Substitute 13 and –13 into the original equation.

9. Example 1B: Using Square Roots to Solve x2 = a
Solve using square roots.
x2 = –49
There is no real number whose square is negative.
There is no real solution.

10. Substitute 11 and –11 into the original equation.
Check It Out! Example 1a
Solve using square roots. Check your answer.
x2 = 121
Solve for x by taking the square root of both sides. Use ± to show both square roots.
x = ± 11
The solutions are 11 and –11.
Check x2 = 121
(11) 
x2 = 121
(–11) 
Substitute 11 and –11 into the original equation.

11. Substitute 0 into the original equation.
Check It Out! Example 1b
Solve using square roots. Check your answer.
x2 = 0
Solve for x by taking the square root of both sides. Use ± to show both square roots.
x = 0
The solution is 0.
Check x2 = 0
(0)2 0 
Substitute 0 into the original equation.

12. Check It Out! Example 1c
Solve using square roots. Check your answer.
x2 = –16
There is no real number whose square is negative.
There is no real solution.

13. If necessary, use inverse operations to isolate the squared part of a quadratic equation before taking the square root of both sides.

14. Example 2A: Using Square Roots to Solve Quadratic Equations
Solve using square roots.
x2 + 7 = 7
–7 –7
x2 + 7 = 7
x2 = 0
Subtract 7 from both sides.
Take the square root of both sides.
The solution is 0.

15. Example 2B: Using Square Roots to Solve Quadratic Equations
Solve using square roots.
16x2 – 49 = 0
16x2 – 49 = 0
Add 49 to both sides.
Divide both sides by 16.
Take the square root of both sides. Use ± to show both square roots.

16.   Example 2B Continued Solve using square roots. 16x2 – 49 = 0
Check x2 – 49 = 0
16x2 – 49 = 0 
49 – 
49 –

17. Check It Out! Example 2a
Solve by using square roots. Check your answer.
100x = 0
100x = 0
–49 –49
100x2 =–49
Subtract 49 from both sides.
Divide both sides by 100.
There is no real number whose square is negative.
There is no real solution.

18. Check It Out! Example 2b
Solve by using square roots. Check your answer.
(x – 5)2 = 16
(x – 5)2 = 16
Take the square root of both sides. Use ± to show both square roots.
x – 5 = ±4
x – 5 = 4 or x – 5 = –4
Write two equations, using both the positive and negative square roots, and solve each equation.
x = 9 or x = 1
The solutions are 9 and 1.

19. Check It Out! Example 2b Continued
Solve by using square roots. Check your answer.
(x – 5)2 = 16
(1 – 5)
(– 4)
Check (x – 5)2 = 16
(9 – 5)  

20. When solving quadratic equations by using square roots, you may need to find the square root of a number that is not a perfect square. In this case, the answer is an irrational number. You can approximate the solutions.

21. Example 3A: Approximating Solutions
Solve. Round to the nearest hundredth.
x2 = 15
Take the square root of both sides.
x  3.87
Evaluate on a calculator.
The approximate solutions are 3.87 and –3.87.

22. Example 3B: Approximating Solutions
Solve. Round to the nearest hundredth.
–3x = 0
–3x = 0
–90 –90
Subtract 90 from both sides.
Divide by – 3 on both sides.
x2 = 30
Take the square root of both sides.
x  5.48
Evaluate on a calculator.
The approximate solutions are 5.48 and –5.48.

23. Example 3B Continued
Solve. Round to the nearest hundredth.
–3x = 0
The approximate solutions are 5.48 and –5.48.
Check Use a graphing calculator to support your answer.
Use the zero function. The approximate solutions are 5.48 and – 5.48.

24. Check It Out! Example 3a
Solve. Round to the nearest hundredth.
0 = 90 – x2
+ x x2
0 = 90 – x2
x2 = 90
Add x2 to both sides.
Take the square root of both sides.
The approximate solutions are 9.49 and –9.49.

25. Check It Out! Example 3b
Solve. Round to the nearest hundredth.
2x2 – 64 = 0
2x2 – 64 = 0
Add 64 to both sides.
Divide by 2 on both sides.
x2 = 32
Take the square root of both sides.
The approximate solutions are 5.66 and –5.66.

26. Check It Out! Example 3c
Solve. Round to the nearest hundredth.
x = 0
x = 0
Subtract 45 from both sides.
– 45 – 45
x2 = –45
There is no real number whose square is negative.
There is no real solution.

27. Let x represent the width of the garden.
Example 4: 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 = A
Use the formula for area of a rectangle.
l = 2w
Length is twice the width.
2x x = 578 
Substitute x for w, 2x for l, and 578 for A.
2x2 = 578

28. Example 4 Continued
2x2 = 578
Divide both sides by 2.
Take the square root of both sides.
x = ± 17
Evaluate on a calculator.
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.

29. Use the formula for area of a trapezoid.
Check It Out! Example 4
A house is on a lot that is shaped like a trapezoid. The solid lines show the boundaries, where x represents the width of the front yard. Find the width of the front yard, given that the area is 6000 square feet. Round to the nearest foot. 2x x
(Hint: Use )
Use the formula for area of a trapezoid.

30. Check It Out! Example 4
Substitute 2x for h and b1, x for b2 , and 6000 for A.
Divide by 3 on both sides.
Take the square root of both sides.
Evaluate on a calculator.
Negative numbers are not reasonable for width, so x ≈ 45 is the only solution that makes sense. Therefore, the width of the front yard is about 45 feet.

31. Solve using square roots. Check your answers. 1. x2 – 195 = 1
Lesson Quiz: Part 1
Solve using square roots. Check your answers.
1. x2 – 195 = 1
2. 4x2 – 18 = –9
3. (x + 7)2 = 81
4. Solve 0 = –5x Round to the nearest hundredth.
± 14
– 16, 2
± 6.71

32. 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