1. Warm Up(Add to HW)
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

2. Solving Quadratic Equations by Using Square Roots 9-7
Holt Algebra 1

3. 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

5.   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.

6. 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.

7. 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.

8. 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.

9. 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 by 16 on both sides.
Take the square root of both sides. Use ± to show both square roots.

10. 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.

11. 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.

12. 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

13. 1) 2) 3) )
5) ) 7)

14. 1) 2) 3) )
5) ) 7)

15. Lesson Quiz: Part 1
Solve using square roots. Check your answers.
1. x2 – 195 = 1
2. 4x2 – 18 = –9
3. 2x2 – 10 = –12
4. Solve 0 = –5x Round to the nearest hundredth.
± 14
no real solutions
± 6.71

16. 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