Warm Up
Find each square root.
1.
6
2.
–25
3.
Solve each equation.
5. –6x = –60
x = 10
7. 2x – 40 = 0
x = 20
11
4.
6.
x = 80
8. 5x = 3
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.
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.)
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
Reading Math
The expression ±3 means “3 or –3” and is read
“plus or minus three.”
Solve using square roots.
x2 = 169
x = ± 13
The solutions are 13 and –13.
Check x2 = 169
(13)2 169
169
169
Solve for x by taking
the square root of
both sides. Use ±
to show both
square roots.
Substitute 13 into
the original
equation.
Solve using square roots.
x2 = –49
There is no real solution. The
solution set is the empty set, ø.
There is no real
number whose
square is
negative.
Solve using square roots
1. x2 = 121
2. x2 = -25
3. x2 = 0
x = ±11
no real solution. ø
x=0
4. x2 = -16
no real solution. ø
5. x2 = 100
x = ±10
6. x2 = 64
x = ±8
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.
Solve using square roots.
x2 + 7 = 32
x2 + 7 = 32
–7 –7
x2 = 25
Subtract 7 from both
sides.
Take the square root
of both sides.
Solve by 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.
Solve by using square roots.
36x2 = 1
Divide by 36 on both
sides.
Take the square root
of both sides. Use
± to show both
square roots.
Solve by using square roots.
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
Solve. Round to the nearest hundredth.
0 = 90 – x2
0 = 90 – x2
+ x2
+ 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.
Solve using square roots
1. x2 -196 = 0
x = ±14
2. 0 = 3x2 -48
x = ±4
3. 24x2 +96 = 0
no real solution. ø
4. 10x2 - 75= 15
x = ±3
5. 0 = 4x2 +144
no real solution. ø
6. 5x2 – 105 = 20
x = ±5
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.
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.
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
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
)