1. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
(For help, go to Lessons 1-2 and 8-5.)
Simplify each expression.
(–12)2 3. –(12) 1 2 3 – 4 5
10-3

2. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Solutions
= 11 • 11 = (–12)2 = (–12)(–12) = 144
3. (–12)2 = –(12)(12) = – = (1.5)(1.5) = 2.25
= (0.6)(0.6) = = • =
= = = • = 1 2 1 2 1 2 1 4 2 3 – 2 3 – 2 3 – 4 9 4 5 4 5 4 5 16 25
10-3

3. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Simplify each expression. a
positive square root
= 5
b. ± 9 25 3 5
The square roots are and – .
= ±
c. – 64
negative square root
= –8
d. –49
For real numbers, the square root of a negative
number is undefined.
is undefined
e. ± 0
There is only one square root of 0.
= 0
10-3

4. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Tell whether each expression is rational or irrational.
a. ± = ± 12
rational
b. – = – 1 5
irrational
c. – = –2.5
rational
d = 0.3 1 9
rational
e =
irrational
10-3

5. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Between what two consecutive integers is ?
28.34 is between the two consecutive perfect squares 25 and 36.
25 < <
The square roots of 25 and 36 are 5 and 6, respectively.
28.34
5 < < 6
28.34 is between 5 and 6.
10-3

6. Finding and Estimating Square Roots
ALGEBRA 1 LESSON 10-3
Find to the nearest hundredth.
28.34
Use a calculator.
5.32
Round to the nearest hundredth.
10-3

7. 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 = (3 • 8)2
Substitute 8 for x.
Simplify.
d =
d =
Use a calculator. Round to the nearest tenth. d
The diagonal is about 25.3 ft long.
10-3

8. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
(For help, go to Lesson10-3.)
Simplify each expression.
– ± ±
± 9. ± 1 4 9 49
100
10-4

9. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solutions
= 6 2. – = –9
3. ± = ± = 1.2
= ± = ±1.1
= 8. ± = ±
9. ± = 1 4 1 2 1 9 1 3 49
100 7 10
10-4

10. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve each equation by graphing the related function.
a. 2x2 = 0
b. 2x2 + 2 = 0
c. 2x2 – 2 = 0
Graph y = 2x2
Graph y = 2x2 + 2
Graph y = 2x2 – 2
There is one solution, x = 0.
There is no solution.
There are two solutions, x = ±1.
10-4

11. Solving Quadratic Equations
ALGEBRA 1 LESSON 10-4
Solve 3x2 – 75 = 0.
3x2 – = Add 75 to each side.
3x2 = 75
x2 = 25 Divide each side by 3.
x = ± Find the square roots.
x = ± 5 Simplify.
10-4

12. 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 r 2
315 = 4 r 2
Substitute 315 for S.
Put in calculator ready form.
315
(4)
= r 2
= r 2
315
(4)
Find the principle square root. r
Use a calculator.
The radius of the sphere is about 5 ft.
10-4

13. 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
b. x2 + 2 = –2
2. Solve each equation by finding square roots.
a. m2 – 25 = 0
b. 49q2 = 9
3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160 joules. Use the equation E = ms2, where m is the object’s mass in kg, E is its kinetic energy, and s is the speed in meters per second. ±2
no solution ±5 3 7 1 2
about 8.94 m/s
10-4

14. 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.
n = – 9 = q + 16 = –3
Factor each expression.
4. 2c2 + 29c p2 + 32p x2 – 21x – 18 a 8
10-5

15. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
Solutions
n = 2
4n = –4
n = –1
Check: 6 + 4(–1) = 6 + (–4) = 2
2. – 9 = 4
= 13
a = 104
Check: – 9 = 13 – 9 = 4
3. 7q + 16 = –3
7q = –19
q = –2
Check: 7 (–2 ) + 16 = 7(– ) + 16 = – = –3 a 8 a 8
104 8 5 7 5 7 19 7
10-5

16. 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

17. 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 = or x – 4 = 0
Use the Zero-Product Property.
2x = –3
Solve for x.
x = – 3 2 or
x = 4
Check: Substitute – for x. 3 2
Substitute 4 for x.
(2x + 3)(x – 4) = 0
[2(– ) + 3](– – 4) 0
[2(4) + 3](4 – 4) 0
(0)(– 5 ) = 0 1
(11)(0) = 0
10-5

18. 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

19. 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
Factor 3x2 – 2x – 21.
3x + 7 = 0 or x – 3 = 0
Use the Zero-Product Property
3x = –7
Solve for x.
x = – or x = 3 7 3
10-5

20. 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 = x + 2
The length of the material = x = x + 5
Relate: length  width = area of the sheet
10-5

21. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
(continued)
Write: (x + 2) (x + 5) = 130
x2 + 7x + 10 = 130
Find the product (x + 2) (x + 5).
x2 + 7x – 120 = 0
Subtract 130 from each side.
(x – 8) (x + 15) = 0
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

22. Factoring to Solve Quadratic Equations
ALGEBRA 1 LESSON 10-5
1. Solve (2x – 3)(x + 2) = 0.
Solve by factoring.
2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25
–2, 3 2 – 2 3 5 2
–1, 6
10-5

23. Completing the Square Find each 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
Factor.
4. b2 + 10b t2 + 14t n2 – 18n + 81
10-6

24. Completing the Square Solutions
ALGEBRA 1 LESSON 10-6
1. (d – 4)2 = d2 – 2d(4) + 42 = d2 – 8d + 16
2. (x + 11)2 = x2 + 2x(11) = 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
Solutions
10-6

25. 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 2. x2 + 7x + c 3. x2 – 9x + c
Solve each equation by completing the square.
4. x2 – 10x + 24 = 0 5. x2 + 16x – 36 = 0
6. 3x2 + 12x – 15 = x2 – 2x – 112 = 0
10-7

26. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions
1. x2 + 6x + c; c = = 32 = x2 + 7x + c; c = =
3. x2 – 9x + c; c = =
4. x2 – 10x + 24 = 0; 5. x2 + 16x – 36 = 0;
c = = (–5)2 = c = = 82 = 64
x2 – 10x = – x2 + 16x = 36
x2 – 10x = – x2 + 16x =
(x – 5)2 = (x + 8)2 = 100
(x – 5) = ± (x + 8) = ±10
x – 5 = 1   or x – 5 = –1 x + 8 = 10  or x + 8 = –10
x = 6   or   x = x = 2   or  x = –18 6 2 7 2 49 4 –9 2 81 4
–10 2 16 2
10-7

27. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solutions (continued)
6. 3x2 + 12x – 15 = x2 – 2x – 112 = 0
3(x2 + 4x – 5) = (x2 – x – 56) = 0
x2 + 4x – 5 = 0; x2 – x – 56 = 0;
c = = 22 = c = =
x2 + 4x = x2 – x = 56
x2 + 4x + 4 = x2 – x = 56 +
(x + 2)2 = (x – )2 =
(x + 2) = ± x – = ±
x + 2 = 3   or x + 2 = –3 x – = or x – =
x = 1   or   x = – x = or x = –7 4 2 –1 2 1 4 1 4 1 4 1 2
225 4 1 2 15 2 1 2 15 2 1 2
–15 2
10-7

28. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve x2 + 2 = –3x using the quadratic formula.
x2 + 3x + 2 = 0
Add 3x to each side and write in standard form.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–3 ± (–3)2 – 4(1)(2)
2(1)
Substitute 1 for a, 3 for b, and 2 for c.
x =
–3 ± 1 2
Simplify.
x =
–3 + 1 2
–3 – 1 or
Write two solutions.
x = –1 or x = –2
Simplify.
10-7

29. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
Check: for x = –1 for x = –2
(–1)2 + 3(–1)
(–2)2 + 3(–2)
1 –
4 –
0 = 0
10-7

30. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.
x =
–b ± b2 – 4ac 2a
Use the quadratic formula.
x =
–4 ± – 4(3)(–8)
2(3)
Substitute 3 for a, 4 for b, and –8 for c.
–4 ± 6
x = or
Write two solutions. – 6
–4 –
Use a calculator. x – 6
–4 – or x
x –2.43
Round to the nearest hundredth.
10-7

31. 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

32. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
(continued)
t =
–15 ± – 4(–16)(2)
2(–16)
Substitute –16 for a, 15 for b, 2 for c, and t for x.
–15 ±
–32
t =
Simplify.
–15 ±
t = –
–32 or
–15 – 18.79
Write two solutions. t
–0.12 or
1.06
Simplify. Use the positive answer because it is the only reasonable answer in this situation.
The ball will land in about 1.06 seconds.
10-7

33. 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 = 0
Quadratic formula; the equation cannot be factored easily and the numbers are large.
10-7

34. Using the Quadratic Formula
ALGEBRA 1 LESSON 10-7
1. Solve 2x2 – 11x + 12 = 0 by using the quadratic formula.
2. Solve 4x2 – 12x = 64. Round the solutions to the nearest hundredth.
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.
1.5, 4
–2.77, 5.77
6.27 seconds
10-7