5. Objectives:
At the end of the lesson, the students
should be able to:
a.find the sum and product of the roots of
quadratic equation;
b.determine the sum and product of the roots
given the roots; and
c.solve quadratic equation in real-life situation
involving the sum and products of the roots.
11. 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.
12. 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
Positive
Square root of 9
15. Example 1A: Using Square Roots to Solve x2 = a
Solve using square roots. Check your answer.
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 and –13
into the original
equation.
x2 = 169
(–13)2 169
169 169
Check x2 = 169
(13)2 169
169 169
16. 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.
17. Check It Out! Example 1a
Solve using square roots. Check your answer.
x2 = 121
x = ± 11
The solutions are 11 and –11.
Solve for x by taking the square root
of both sides. Use ± to show both
square roots.
Substitute 11 and –11
into the original
equation.
x2 = 121
(–11)2 121
121 121
Check x2 = 121
(11)2 121
121 121
18. Check It Out! Example 1b
Solve using square roots. Check your answer.
x2 = 0
x = 0
The solution is 0.
Solve for x by taking the square root
of both sides. Use ± to show both
square roots.
Substitute 0 into the
original equation.
Check x2 = 0
(0)2 0
0 0
19. x2 = –16
There is no real number whose
square is negative.
There is no real solution.
Check It Out! Example 1c
Solve using square roots. Check your answer.
20. If necessary, use inverse
operations to isolate the squared
part of a quadratic equation
before taking the square root of
both sides.
21. Example 2A: Using Square Roots to Solve Quadratic
Equations
Solve using square roots.
x2 + 7 = 7
–7 –7
x2 + 7 = 7
x2 = 0
The solution is 0.
Subtract 7 from both sides.
Take the square root of both
sides.
22. Example 2B: Using Square Roots to Solve Quadratic
Equations
Solve using square roots.
16x2 – 49 = 0
16x2 – 49 = 0
+49 +49 Add 49 to both sides.
Divide both sides by 16.
Take the square root of both
sides. Use ± to show both
square roots.
23. Example 2B Continued
Solve using square roots.
Check 16x2 – 49 = 0
49 – 49 0
16x2 – 49 = 0
49 – 49 0
.
24. Check It Out! Example 2a
Solve by using square roots. Check your answer.
100x2 + 49 = 0
100x2 + 49 = 0
–49 –49
100x2 =–49
There is no real solution.
There is no real number
whose square is negative.
Subtract 49 from both sides.
Divide both sides by 100.
25. Check It Out! Example 2b
Solve by using square roots. Check your answer.
(x – 5)2 = 16
Take the square root of both sides.
Use ± to show both square roots.
(x – 5)2 = 16
x – 5 = ±4
Write two equations, using both the
positive and negative square roots,
and solve each equation.
x – 5 = 4 or x – 5 = –4
+ 5 + 5 + 5 + 5
x = 9 or x = 1
The solutions are 9 and 1.
26. Check It Out! Example 2b Continued
Solve by using square roots. Check your answer.
Check (x – 5)2 = 16
(9 – 5)2 16
42 16
16 16
(x – 5)2 = 16
(1 – 5)2 16
(– 4)2 16
16 16
27. 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.
28. Example 3A: Approximating Solutions
Solve. Round to the nearest hundredth.
x2 = 15
Take the square root of both sides.
Evaluate on a calculator.
The approximate solutions are 3.87 and –3.87.
x 3.87
29. Example 3B: Approximating Solutions
Solve. Round to the nearest hundredth.
–3x2 + 90 = 0
–3x2 + 90 = 0
–90 –90
x2 = 30
The approximate solutions are 5.48 and –5.48.
Subtract 90 from both sides.
Divide by – 3 on both sides.
Take the square root of both
sides.
Evaluate on a calculator.
x 5.48
30. Check It Out! Example 3b
Solve. Round to the nearest hundredth.
2x2 – 64 = 0
2x2 – 64 = 0
+ 64 + 64
x2 = 32
The approximate solutions are 5.66 and –5.66.
Add 64 to both sides.
Divide by 2 on both sides.
Take the square root of both
sides.
31. Check It Out! Example 3c
Solve. Round to the nearest hundredth.
x2 + 45 = 0
x2 + 45 = 0
– 45 – 45
x2 = –45
There is no real number whose
square is negative.
There is no real solution.
Subtract 45 from both sides.
32. 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.
Substitute x for w, 2x for l, and
578 for A.
2x x = 578
l = 2w
2x2 = 578
Length is twice the width.
33. Example 4 Continued
2x2 = 578
x = ± 17
Take the square root of both sides.
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.
Divide both sides by 2.
34. 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.
35. 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 = –5x2 + 225. Round to the nearest
hundredth.
± 14
± 6.71
– 16, 2
36. 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
