Finding and Estimating Square Roots (For help, go to Lessons 1-2 and 8-5.) ALGEBRA 1 LESSON 10-3 Simplify each expression. 1.11 2 2.(–12) 2 3.–(12) 2 4.1.5

Finding and Estimating Square RootsFinding and Estimating Square Roots

(For help, go to Lessons 1-2 and 8-5.)

ALGEBRA 1 LESSON 10-3ALGEBRA 1 LESSON 10-3

Simplify each expression.

1. 112 2. (–12)2 3. –(12)2 4. 1.52

5. 0.62 6. 2 7. 2 8. 212

23–

45

10-3

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. 2 = • =

7. 2 = = 8.

2 = • =

Solutions

12

23– 4

5

12

12

14

23–

23– 4

545

1625

Finding and Estimating Square RootsFinding and Estimating Square RootsALGEBRA 1 LESSON 10-3ALGEBRA 1 LESSON 10-3

10-3

49

a. 25



c. – 64

d. –49

e. ± 0

b. ± 925

10-3

positive square root= 5

35

35

35

The square roots are and – .= ±

negative square root= –8

For real numbers, the square root of a negative number is undefined.

is undefined

There is only one square root of 0. = 0

a. ± 144 = ± 12


Tell whether each expression is rational or irrational.

c. – 6.25 = –2.5

e. 7 = 2.64575131 . . .

b. – = –0.44721359 . . .15

d. = 0.319

10-3

rational

irrational

rational

rational

irrational

Between what two consecutive integers is 28.34 ?


28.34 is between 5 and 6.

28.34 is between the two consecutive perfect squares 25 and 36.

25 < 28.34 < 36

The square roots of 25 and 36 are 5 and 6, respectively.

28.345 < < 6

10-3

Find 28.34 to the nearest hundredth.


28.34 5.323532662 Use a calculator.

5.32 Round to the nearest hundredth.

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.

The diagonal is about 25.3 ft long.

Use a calculator. Round to the nearest tenth.d 25.3

10-3

Simplify.d = 64 + 576

d = 640

Solving Quadratic EquationsSolving Quadratic EquationsALGEBRA 1 LESSON 10-4ALGEBRA 1 LESSON 10-4

(For help, go to Lesson10-3.)


1. 36 2. – 81 3. ± 121

4. 1.44 5. 0.25 6. ± 1.21

7. 8. ± 9. ±14

19

49100

10-4


Solutions

13

14

19

49100

12

710

10-4

1. 36 = 6 2. – 81 = –9

3. ± 121 = ±11 4. 1.44 = 1.2

5. 0.25 = 0.5 6. ± 1.21 = ±1.1

7. = 8. ± = ±

9. ± =


Solve each equation by graphing the related function.

a. 2x2 = 0 b. 2x2 + 2 = 0 c. 2x2 – 2 = 0

There is one solution, x = 0.

There is no solution. There are two solutions, x = ±1.

10-4

Graph y = 2x2 Graph y = 2x2 + 2 Graph y = 2x2 – 2

x = ± 25 Find the square roots.


Solve 3x2 – 75 = 0.

3x2 – 75 + 75 = 0 + 75 Add 75 to each side.

3x2 = 75

x2 = 25 Divide each side by 3.

x = ± 5 Simplify.

10-4


S = 4 r 2

315 = 4 r 2 Substitute 315 for S.

Put in calculator ready form.315 (4)

= r 2

= r 2315 (4)

Find the principle square root.

5.00668588 r Use a calculator.

The radius of the sphere is about 5 ft.

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.


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

37±

12

about 8.94 m/s

10-4

Factoring to Solve Quadratic EquationsFactoring to Solve Quadratic EquationsALGEBRA 1 LESSON 10-5ALGEBRA 1 LESSON 10-5


Solve and check each equation.

1. 6 + 4n = 2 2. – 9 = 4 3. 7q + 16 = –3

Factor each expression.

4. 2c2 + 29c + 14 5. 3p2 + 32p + 20 6. 4x2 – 21x – 18

a8

10-5


Solutions

1. 6 + 4n = 2 4n = –4

n = –1Check: 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 = –19 + 16 = –3

a8

a8

1048

57

57

197

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


Solve (2x + 3)(x – 4) = 0 by using the Zero Product Property.

(2x + 3)(x – 4) = 0

2x + 3 = 0 or x – 4 = 0 Use the Zero-Product Property.

2x = –3 Solve for x.

x = – 32 or x = 4

Check: Substitute – for x.32 Substitute 4 for x.

(2x + 3)(x – 4) = 0 (2x + 3)(x – 4) = 0

[2(– ) + 3](– – 4) 032

32 [2(4) + 3](4 – 4) 0

(0)(– 5 ) = 012 (11)(0) = 0

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


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 = 373

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 + 2The length of the material = x + 3 + 1 + 1 = x + 5

Relate: length width = area of the sheet

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


1. Solve (2x – 3)(x + 2) = 0.

Solve by factoring.

2. 6 = a2 – 5a 3. 12x + 4 = –9x2 4. 4y2 = 25

–2, 32

–1, 6 –23

±52

10-5

Completing the SquareCompleting the SquareALGEBRA 1 LESSON 10-6ALGEBRA 1 LESSON 10-6


Find each square.

1. (d – 4)2 2. (x + 11)2 3. (k – 8)2

Factor.

4. b2 + 10b + 25 5. t2 + 14t + 49 6. n2 – 18n + 81

10-6

Completing the SquareCompleting the SquareALGEBRA 1 LESSON 10-6ALGEBRA 1 LESSON 10-6

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

Solutions

10-6

Using the Quadratic FormulaUsing the Quadratic Formula

(For help, go to Lesson 10-6.)

ALGEBRA 1 LESSON 10-7ALGEBRA 1 LESSON 10-7

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 = 0 7. 2x2 – 2x – 112 = 0

10-7

Using the Quadratic FormulaUsing the Quadratic FormulaALGEBRA 1 LESSON 10-7ALGEBRA 1 LESSON 10-7

1. x2 + 6x + c; c = 2 = 32 = 9 2. x2 + 7x + c; c = 2 =

3. x2 – 9x + c; c = 2 =

4. x2 – 10x + 24 = 0;5. x2 + 16x – 36 = 0;

c = 2 = (–5)2 = 25 c = 2 = 82 = 64

x2 – 10x = –24 x2 + 16x = 36x2 – 10x + 25 = –24 + 25 x2 + 16x + 64 = 36 + 64 (x – 5)2 = 1 (x + 8)2 = 100 (x – 5) = ±1 (x + 8) = ±10x – 5 = 1 or x – 5 = –1 x + 8 = 10 or x + 8 = –10 x = 6 or x = 4 x = 2 or x = –18

62

72

494

–92

814

–102

162

Solutions

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;

c = 2 = 22 = 4 c = 2 =

x2 + 4x = 5 x2 – x = 56

x2 + 4x + 4 = 5 + 4 x2 – x + = 56 +

(x + 2)2 = 9 (x – )2 =

(x + 2) = ±3 x – = ±

x + 2 = 3 or x + 2 = –3 x – = or x – =

x = 1 or x = –5 x = 8 or x = –7

42

–12

14

14

14

12

2254

12

152

12

152

12

–152

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 ± 12

Simplify.

x =–3 + 1

2 x =–3 – 1

2or Write two solutions.

x = –1 or x = –2 Simplify.

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


Solve 3x2 + 4x – 8 = 0. Round the solutions to the nearest hundredth.

x =–b ± b2 – 4ac

2a Use the quadratic formula.

x = –4 ± 42 – 4(3)(–8)2(3)

Substitute 3 for a, 4 for b, and –8 for c.

–4 ± 1126

x =

Use a calculator.x–4 + 10.583005244

6 x–4 – 10.583005244

6or

x 1.10 or x –2.43Round to the nearest hundredth.

10-7

x = or x = Write two solutions.–4 + 1126

–4 – 1126


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 + c0 = –16t2 + 15t + 2 Substitute 0 for h, 15 for v, and 2 for c.

Step 2: Use the quadratic formula.

x = –b ± b2 – 4ac2a

10-7


(continued)

t = –15 ± 152 – 4(–16)(2)2(–16)

Substitute –16 for a, 15 for b, 2 for c, and t for x.

t =–15 + 18.79

–32 or t =–15 – 18.79

–32 Write two solutions.

t –0.12 or t 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.

–15 ± 225 + 128–32t = Simplify.

–15 ± 353–32t =

10-7


Which method(s) would you choose to solve each equation?

Justify your reasoning.

10-7

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.


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

Documents

Finding and Estimating Square Roots (For help, go to Lessons 1-2 and 8-5.) ALGEBRA 1 LESSON 10-3 Simplify each expression. 1.11 2 2.(–12) 2 3.–(12) 2 4.1.5