Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
(For help, go to Lessons 1-2 and 8-5.)
Simplify each expression.
1. 112
2. (–12)2
5. 0.62
6. 1
2
3. –(12)2
7.
2
10-3
2
–3
2
4. 1.52
8. 4
5
2
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Solutions
1. 112 = 11 • 11 = 121
2. (–12)2 = (–12)(–12) = 144
3. (–12)2 = –(12)(12) = –144
4. 1.52 = (1.5)(1.5) = 2.25
5. 0.62 = (0.6)(0.6) = 0.36
6. 1
7. – 2
3
2
= –2
3
2
–3
2
2
2
= 4
8. 4
9
5
10-3
= 1 • 1 = 1
2
2
4
= 4 • 4 = 16
5
5
25
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Simplify each expression.
a.
25 = 5
positive square root
b. ±
3
9
=± 5
25
The square roots are
c. –
64 = –8
negative square root
d.
e. ±
–49 is undefined
0 =0
3
3
and – .
5
5
For real numbers, the square root of a negative
number is undefined.
There is only one square root of 0.
10-3
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Tell whether each expression is rational or irrational.
a. ±
144 = ± 12
rational
b. –
1
= –0.44721359 . . .
5
irrational
c. –
6.25 = –2.5
rational
d.
1
= 0.3
9
rational
e.
7 = 2.64575131 . . .
irrational
10-3
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Between what two consecutive integers is
25 <
5<
28.34 <
36
28.34 < 6
28.34 ?
28.34 is between the two consecutive
perfect squares 25 and 36.
The square roots of 25 and 36 are 5 and
6, respectively.
28.34 is between 5 and 6.
10-3
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Find
28.34
28.34 to the nearest hundredth.
5.323532662
Use a calculator.
5.32
Round to the nearest hundredth.
10-3
Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Suppose a rectangular field has a length three times its width
x. The formula d = x2 + (3x)2 gives the distance of the diagonal of a
rectangle. Find the distance of the diagonal across the field if x = 8 ft.
d=
x2 + (3x)2
d=
82 + (3 • 8)2
Substitute 8 for x.
d=
64 + 576
Simplify.
d=
640
d
25.3
Use a calculator. Round to the nearest tenth.
The diagonal is about 25.3 ft long.
10-3
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
(For help, go to Lesson10-3.)
Simplify each expression.
1.
36
2. –
4.
1.44
5.
7.
1
4
8. ±
81
3. ±
121
0.25
6. ±
1.21
9. ±
49
100
1
9
10-4
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solutions
1.
3. ±
36 = 6
121 = ±11
2. –
4.
81 = –9
1.44 = 1.2
5.
0.25 = 0.5
6. ±
1.21 = ±1.1
7.
1
= 1
4
2
8. ±
1
1
=±
3
9
9. ±
49 = 7
10
100
10-4
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve each equation by graphing the related function.
a. 2x2 = 0
Graph y = 2x2
There is one
solution, x = 0.
b. 2x2 + 2 = 0
Graph y = 2x2 + 2
There is no solution.
10-4
c. 2x2 – 2 = 0
Graph y = 2x2 – 2
There are two
solutions, x = ±1.
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve 3x2 – 75 = 0.
3x2 – 75 + 75 = 0 + 75
Add 75 to each side.
3x2 = 75
x2 = 25
x=±
x=±5
Divide each side by 3.
25
Find the square roots.
Simplify.
10-4
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
A museum is planning an exhibit that will contain a large globe.
The surface area of the globe will be 315 ft2. Find the radius of the sphere
producing this surface area. Use the equation S = 4 r2, where S is the
surface area and r is the radius.
S = 4 r2
315 = 4 r 2
Substitute 315 for S.
315
= r2
(4)
Put in calculator ready form.
315
= r2
(4)
Find the principle square root.
5.00668588
r
Use a calculator.
The radius of the sphere is about 5 ft.
10-4
Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
1. Solve each equation by graphing the related function. If the equation
has no solution, write no solution.
a. 2x2 – 8 = 0
±2
b. x2 + 2 = –2 no solution
2. Solve each equation by finding square roots.
a. m2 – 25 = 0 ±5
b. 49q2 = 9
3
±7
3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160
joules. Use the equation E = 1 ms2, where m is the object’s mass in kg,
2
E is its kinetic energy, and s is the speed in meters per second.
about 8.94 m/s
10-4
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(For help, go to Lessons 2-2 and 9-6.)
Solve and check each equation.
1. 6 + 4n = 2
2. a – 9 = 4
3. 7q + 16 = –3
5. 3p2 + 32p + 20
6. 4x2 – 21x – 18
8
Factor each expression.
4. 2c2 + 29c + 14
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solutions
1. 6 + 4n = 2
4n = –4
n = –1
Check: 6 + 4(–1) = 6 + (–4) = 2
a
2. 8 – 9 = 4
a
= 13
8
a = 104
Check: 104 – 9 = 13 – 9 = 4
8
3. 7q + 16 = –3
7q = –19
q = –2 5
7
5
Check: 7 (–2 ) + 16 = 7(– 19 ) + 16 = –19 + 16 = –3
7
7
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solutions (continued)
4. 2c2 + 29c + 14 = (2c + 1)(c + 14)
Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14
5. 3p2 + 32p + 20 = (3p + 2)(p + 10)
Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20
6. 4x2 – 21x – 18 = (4x + 3)(x – 6)
Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.
(2x + 3)(x – 4) = 0
2x + 3 = 0
x–4=0
or
2x = –3
3
x=–2
Use the Zero-Product Property.
Solve for x.
or
x=4
3
Check: Substitute – 2 for x.
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(– 3 ) + 3](– 3 – 4)
2
2
(2x + 3)(x – 4) = 0
[2(4) + 3](4 – 4)
0
1
(0)(– 5 2 ) = 0
0
(11)(0) = 0
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve x2 + x – 42 = 0 by factoring.
x2 + x – 42 = 0
(x + 7)(x – 6) = 0
Factor using x2 + x – 42
x+7=0
or
x–6=0
Use the Zero-Product Property.
x = –7
or
x=6
Solve for x.
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solve 3x2 – 2x = 21 by factoring.
3x2 – 2x = 21
Subtract 21 from each side.
(3x + 7)(x – 3) = 0
3x + 7 = 0
Factor 3x2 – 2x – 21.
x–3=0
or
3x = –7
7
x=– 3
Use the Zero-Product Property
Solve for x.
or
x=3
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
The diagram shows a pattern for an open-top box. The total
area of the sheet of materials used to make the box is 130 in.2. The
height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from
each corner. Find the dimensions of the box.
Define: Let x = width of a side of the box.
Then the width of the material = x + 1 + 1 = x + 2
The length of the material = x + 3 + 1 + 1 = x + 5
Relate: length  width = area of the sheet
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(continued)
Write: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130
x2 + 7x – 120 = 0
(x – 8) (x + 15) = 0
Find the product (x + 2) (x + 5).
Subtract 130 from each side.
Factor x2 + 7x – 120.
x–8=0
or
x + 15 = 0
Use the Zero-Product Property.
x=8
or
x = –15
Solve for x.
The only reasonable solution is 8. So the dimensions of the box are
8 in.  11 in.  1 in.
10-5
Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
1. Solve (2x – 3)(x + 2) = 0.
3
–2, 2
Solve by factoring.
2. 6 = a2 – 5a
–1, 6
3. 12x + 4 = –9x2
4. 4y2 = 25
2
3
±
–
10-5
5
2
Completing the Square
ALGEBRA 1 LESSON 10-6
(For help, go to Lessons 9-4 and 9-7.)
Find each square.
1. (d – 4)2
2. (x + 11)2
3. (k – 8)2
5. t2 + 14t + 49
6. n2 – 18n + 81
Factor.
4. b2 + 10b + 25
10-6
Completing the Square
ALGEBRA 1 LESSON 10-6
Solutions
1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16
2. (x + 11)2 = x2 + 2x(11) + 112 = x2 + 22x + 121
3. (k – 8)2 = k2 – 2k(8) + 82 = k2 – 16k + 64
4. b2 + 10b + 25 = b2 + 2b(5) + 52 = (b + 5)2
5. t2 + 14t + 49 = t2 + 2t(7) + 72 = (t + 7)2
6. n2 – 18n + 81 = n2 – 2n(9) + 92 = (n – 9)2
10-6
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(For help, go to Lesson 10-6.)
Find the value of c to complete the square for each expression.
1. x2 + 6x + c
3. x2 – 9x + c
2. x2 + 7x + c
Solve each equation by completing the square.
4. x2 – 10x + 24 = 0
5.
x2 + 16x – 36 = 0
6. 3x2 + 12x – 15 = 0
7.
2x2 – 2x – 112 = 0
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions
1. x2 + 6x + c; c = 6
2
2
3. x2 – 9x + c; c =
–9
2
2
= 32 = 9
2
2
=
81
5. x2 + 16x – 36 = 0;
c = 16
= (–5)2 = 25
– 10x
x2 – 10x + 25
(x – 5)2
(x – 5)
x – 5 = 1 or
x = 6 or
x2
2
= 4
4. x2 – 10x + 24 = 0;
c = –10
7
2. x2 + 7x + c; c = 2
= –24
= –24 + 25
=1
= ±1
x – 5 = –1
x=4
2
2
= 82 = 64
x2 + 16x = 36
x2 + 16x + 64 = 36 + 64
(x + 8)2 = 100
(x + 8) = ±10
x + 8 = 10 or x + 8 = –10
x = 2 or
x = –18
10-7
49
4
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions (continued)
6. 3x2 + 12x – 15 = 0
7. 2x2 – 2x – 112 = 0
3(x2 + 4x – 5) = 0
2(x2 – x – 56) = 0
x2 + 4x – 5 = 0;
x2 – x – 56 = 0;
4
c= 2
2
c = –1
= 22 = 4
2
2
= 1
4
x2 + 4x = 5
x2 – x = 56
x2 + 4x + 4 = 5 + 4
x2 – x + 1 = 56 + 1
(x + 2)2 = 9
(x – 1 )2 = 225
(x + 2) = ±3
x + 2 = 3 or x + 2 = –3
x = 1 or
x = –5
4
2
4
4
x – 1 = ± 15
2
2
1
15
1
x– =
or x – = –15
2
2
2
2
x=8
10-7
or x = –7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve x2 + 2 = –3x using the quadratic formula.
x2 + 3x + 2 = 0
x=
x=
–b ±
–3 ±
b2 – 4ac
2a
Use the quadratic formula.
(–3)2 – 4(1)(2)
2(1)
x=
–3 ±
2
x=
–3 + 1
2
x = –1
Add 3x to each side and write in standard form.
1
Substitute 1 for a, 3 for b, and 2 for c.
Simplify.
–3 – 1
2
or
x=
or
x = –2
Write two solutions.
Simplify.
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
Check:
for x = –1
for x = –2
(–1)2 + 3(–1) + 2
0
(–2)2 + 3(–2) + 2
0
1–3+2
0
4–6+2
0
0=0
0=0
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest
hundredth.
x=
x=
x=
x=
–b ±
b2 – 4ac
2a
–4 ±
42 – 4(3)(–8)
2(3)
–4 ±
112
Use the quadratic formula.
Substitute 3 for a, 4 for b, and –8 for c.
6
–4 +
112
6
or
x=
–4 –
112
6
Write two solutions.
x
–4 + 10.583005244
6
or
x
–4 – 10.583005244
6
Use a calculator.
x
1.10
or
x
–2.43
Round to the
nearest hundredth.
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
A child throws a ball upward with an initial upward velocity of
15 ft/s from a height of 2 ft. If no one catches the ball, after how many
seconds will it land? Use the vertical motion formula h = –16t2 + vt + c,
where h = 0, v = velocity, c = starting height, and t = time to land. Round
to the nearest hundredth of a second.
Step 1: Use the vertical motion formula.
h = –16t2 + vt + c
0 = –16t2 + 15t + 2 Substitute 0 for h, 15 for v, and 2 for c.
Step 2: Use the quadratic formula.
x=
–b ±
b2 – 4ac
2a
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
t=
t=
–15 ±
–15 ±
152 – 4(–16)(2)
2(–16)
225 + 128
–32
t=
–15 ± 353
–32
t=
–15 + 18.79
–32
t
–0.12
Substitute –16 for a, 15 for b, 2 for c, and t for x.
Simplify.
or
t=
or
t
–15 – 18.79
–32
1.06
The ball will land in about 1.06 seconds.
10-7
Write two solutions.
Simplify. Use the positive answer
because it is the only reasonable
answer in this situation.
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Which method(s) would you choose to solve each equation?
Justify your reasoning.
a. 5x2 + 8x – 14 = 0
Quadratic formula; the equation cannot be
factored easily.
b. 25x2 – 169 = 0
Square roots; there is no x term.
c. x2 – 2x – 3 = 0
Factoring; the equation is easily factorable.
d. x2 – 5x + 3 = 0
Quadratic formula, completing the square, or
graphing; the x2 term is 1, but the equation is
not factorable.
e. 16x2 – 96x + 135 = 0
Quadratic formula; the equation cannot be
factored easily and the numbers are large.
10-7
Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.
1.5, 4
2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.
–2.77, 5.77
3. Suppose a model rocket is launched from a platform 2 ft above the
ground with an initial upward velocity of 100 ft/s. After how many
seconds will the rocket hit the ground? Round the solution to the
nearest hundredth.
6.27 seconds
10-7