Chapter 9 Section 6

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9-6
Solving Quadratic Equations by Using Square Roots
Preview
Warm Up
California Standards
Lesson Presentation
9-6
Solving Quadratic Equations by Using Square Roots
Warm Up
Find each square root.
1.
6
2.
11
4.
–25
3.
Solve each equation.
5. –6x = –60
x = 10
7. 2x – 40 = 0
x = 20
6.
8. 5x = 3
x = 80
9-6
Solving Quadratic Equations by Using Square Roots
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.
9-6
Solving Quadratic Equations by Using Square Roots
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.)
9-6
Solving Quadratic Equations by Using Square Roots
Positive
square root of 9
Negative
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 ±√ .
Positive and negative
square roots of 9
9-6
Solving Quadratic Equations by Using Square Roots
Reading Math
The expression ±3 means “3 or –3” and is read
“plus or minus three.”
9-6
Solving Quadratic Equations by Using Square Roots
9-6
Solving Quadratic Equations by Using Square Roots
Additional 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)2
169
169
169
Substitute 13 into the
original equation.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 1A Continued
Solve using square roots. Check your answer.
Check x2 = 169
(–13)2 169
169
169
Substitute –13 into the
original equation.
9-6
Solving Quadratic Equations by Using Square Roots
Additional 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. The solution set is the
empty set, ø.
9-6
Solving Quadratic Equations by Using Square Roots
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
x = ±11
square roots.
The solutions are 11 and –11.
Check x2 = 121
Substitute 11 into the
(11)2 121
original equation.
121
121
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 1a Continued
Solve using square roots. Check your answer.
Check x2 = 121
(–11)2 121
121
121
Substitute –11 into the
original equation.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 1b
Solve using square roots. Check your answer.
x2 = 0
x=0
Solve for x by taking the square root
of both sides. Use ± to show both
square roots.
The solution is 0.
Check x2 = 0
(0)2 0
0
0
Substitute 0 into the
original equation.
9-6
Solving Quadratic Equations by Using Square Roots
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. The solution set is the
empty set, ø.
9-6
Solving Quadratic Equations by Using Square Roots
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.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 2A: Using Square Roots to
Solve Quadratic Equations
Solve using square roots.
x2 + 7 = 7
x2 + 7 = 7
–7 –7
x2 = 0
Subtract 7 from both sides.
Take the square root of both
sides.
9-6
Solving Quadratic Equations by Using Square Roots
Additional 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 by 16 on both sides.
Take the square root of both
sides. Use ± to show both
square roots.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 2B Continued
Solve using square roots.
Check
16x2 – 49 = 0
0
0
16x2 – 49 = 0
0
0
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 2a
Solve by using square roots. Check your answer.
100x2 + 49 = 0
100x2 + 49 = 0
–49 –49
100x2 =–49
Subtract 49 from both sides.
Divide by 100 on both sides.
There is no real number
whose square is negative.
ø
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 2b
Solve by using square roots. Check your answer.
36x2 = 1
Divide by 36 on both sides.
Take the square root of both
sides. Use ± to show both
square roots.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 2b Continued
Solve by using square roots. Check your answer.
Check
36x2 = 1
1
1
36x2 = 1
1
1
9-6
Solving Quadratic Equations by Using Square Roots
When solving quadratic equations by using
square roots, the solutions may be
irrational. In this case, you can give the
exact solutions by leaving the square root
in your answer, or you can approximate the
solutions.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 3A: Approximating Solutions
Solve. Round to the nearest hundredth.
x2 = 15
Take the square root of both sides.
x  3.87
Estimate
.
The exact solutions are
and
.
The approximate solutions are 3.87 and –3.87.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 3B: Approximating Solutions
Solve. Round to the nearest hundredth.
–3x2 + 90 = 0
–3x2 + 90 = 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
Estimate
.
The exact solutions are
and
.
The approximate solutions are 5.48 and –5.48.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 3B Continued
Solve. Round to the nearest hundredth.
–3x2 + 90 = 0
Check Use a graphing calculator to support your
answer.
Use the zero function. The
approximate solutions are
5.48 and –5.48.
9-6
Solving Quadratic Equations by Using Square Roots
Remember!
To review estimating square roots, see Lesson
1-5.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 3a
Solve. Round to the nearest hundredth.
0 = 90 – x2
Add x2 to both sides.
0 = 90 – x2
+ x2
+ x2
x2 = 90
Take the square root of both
sides.
Estimate
The exact solutions are
and
.
The approximate solutions are 9.49 and –9.49.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 3a Continued
Solve. Round to the nearest hundredth.
0 = 90 – x2
Check Use a graphing calculator to support your
answer.
Use the zero function. The
approximate solutions are
9.49 and –9.49.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 3b
Solve. Round to the nearest hundredth.
2x2 – 64 = 0
2x2 – 64 = 0
+ 64 + 64
Add 64 to both sides.
Divide by 2 on both sides.
x2 = 32
Take the square root of both
sides.
Estimate
The exact solutions are
and
. The
approximate solutions are 5.66 and –5.66.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 3b Continued
Solve. Round to the nearest hundredth.
2x2 – 64 = 0
Check Use a graphing calculator to support your
answer.
Use the zero function. The
approximate solutions are
5.66 and –5.66.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 3c
Solve. Round to the nearest hundredth.
x2 + 45 = 0
x2 + 45 = 0
– 45 – 45
x2 = –45
Subtract 45 from both sides.
There is no real number whose
square is negative.
ø
9-6
Solving Quadratic Equations by Using Square Roots
Additional 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
l = 2w
2x
●
x = 578
2x2 = 578
Use the formula for area of a rectangle.
Length is twice the width.
Substitute x for w, 2x for l, and
578 for A.
9-6
Solving Quadratic Equations by Using Square Roots
Additional Example 4 Continued
2x2 = 578
Divide both sides by 2.
Take the square root of both sides.
x = ±17
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.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 4
A lot is shaped like a trapezoid
with bases x and 2x. Its area is
6000 ft2. Find x. Round to the
nearest foot.
(Hint: Use
2x
2x
)
x
Use the formula for area of a trapezoid.
Substitute 2x for h and b1, x
for b2 , and 6000 for A.
9-6
Solving Quadratic Equations by Using Square Roots
Check It Out! Example 4 Continued
Divide by 3 on both sides.
Take the square root of both
sides.
Estimate
.
Negative numbers are not reasonable for width,
so x ≈ 45 is the only solution that makes sense.
Therefore, x is approximately 45 feet.
9-6
Solving Quadratic Equations by Using Square Roots
Lesson Quiz: Part I
Solve using square roots. Check your answers.
1. x2 – 195 = 1
± 14
2. 4x2 – 18 = –9
3. 2x2 – 10 = –12 ø
4. Solve 0 = –5x2 + 225. Round to the nearest
hundredth. ± 6.71
9-6
Solving Quadratic Equations by Using Square Roots
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
)
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