Pre-Calculus 11 Chapter 3 Review: Solving Quadratic Equations

Pre-Calculus 11 Chapter 3 Review: Solving Quadratic Equations Factor each expression.

1) 𝑥 ^" + 8𝑥 ^% − 20𝑥 1_______________________

2) 4𝑥 ^* − 4 2_______________________

3) 4 − (𝑥 + 7) ^% 3)______________________

4) 𝑥 ^" (𝑎 + 𝑏) − 6𝑥 ^% (𝑎 + 𝑏) + 8𝑥(𝑎 + 𝑏) 4)______________________

5) −8𝑥 ^" + 10𝑥 ^% + 12𝑥 5)______________________

6) 25𝑥 ^% (𝑎 − 1) ^" − 5𝑥(𝑎 − 1) ^" − 2(𝑎 − 1) ^" 6)______________________

Solve by factoring.

7) ³

345 − ^"

367 = ^"9

3 ^: 4;345 7)______________________

8) 6𝑥 ^% (3𝑥 − 1) − 𝑥(3𝑥 − 1) = 2(3𝑥 − 1) 8)______________________

9) Given the solutions to a quadratic equation are 𝑥 = ^%±>√@ _% , write the equation in 𝑎𝑥 ^% + 𝑏𝑥 + 𝑐 = 0

9)______________________

Solve by using the square root property/completing the square.

10) 3𝑥 ^% + 18𝑥 − 24 = 0 10)_____________________

11) −𝑥 ^% + 8𝑥 − 17 = 0 11)_____________________

12) 2𝑥 ^% − 2√2𝑥 = −1 12)_____________________

13) Identify the values of a, b, c. 6𝑥(𝑥 + 1) = (𝑥 + 3) ^% 13)_____________________

14) The length of a rectangle is three cm less than twice the width. Find the dimensions of the rectangle if the area is 20 cm ² .

14)_____________________

𝑥 = −𝑏 ± √𝑏 ^% − 4𝑎𝑐 2𝑎

15) Solve √𝑥 + 3 = 2𝑥 − 1 by using the quadratic formula. (Exact Value) 15)_____________________

16) Solve ⁷ ₃ − ^%

34; = 2 by using the quadratic formula. (Three Decimal Places) 16)_____________________

17) Find the values of k so that 𝑥 ^% + 𝑘𝑥 + 3𝑘 − 5 = 0 will have two identical real roots.

17)_____________________

18) Find the values of k so that 𝑥 ^% − 𝑘𝑥 + 9 = 0 will have two real and distinct roots.

18)_____________________

19) Solve −2𝑥 ^% + 12𝑥 − 10 = 0 by graphing. 19)_____________________

20) Solve 0.25𝑥 ^% + 1.5𝑥 − 1.75 = 0 by graphing. 20)_____________________

21) A 20 cm by 60 cm painting has a frame surrounding it. If the frame is the same width all around, and the total area of the frame is 516 cm ² , how wide is the frame?

21)_____________________

22) The total surface area of a square base box with a height of 9 cm is 350 cm ² . Find the length of the base of the box.

22)_____________________

23) Pedro drives 300 km. If he increases his speed by 10 km/h, it takes 1 hour less time. What was Pedro's original speed?

23)_____________________

24) Two pipes together can fill a small reservoir in 2 hours. One of the pipes used alone takes 3 hours longer than the other to fill the reservoir. How long does it take the slower pipe to fill the reservoir alone?

25)_____________________

25) A circular lawn is surrounded by a flower bed of uniform width. If the flower bed has an area of 36 m ² and the radius of the entire garden is 8 m, find the width of the flower bed.

ANSWER KEY:

1) 𝑥 ^" + 8𝑥 ^% − 20𝑥 2) 4𝑥 ^* − 4

𝑥(𝑥 ^% + 8𝑥 − 20) 4(𝑥 ^* − 1)

𝑥(𝑥 + 10)(𝑥 − 2) 4(𝑥 ^; + 1)(𝑥 ^; − 1)

4(𝑥 ^; + 1)(𝑥 ^% + 1)(𝑥 ^% − 1) 4(𝑥 ^; + 1)(𝑥 ^% + 1)(𝑥 + 1)(𝑥 − 1)

3) 4 − (𝑥 + 7) ^% let (𝑥 + 7) = 𝑘 4) 𝑥 ^" (𝑎 + 𝑏) − 6𝑥 ^% (𝑎 + 𝑏) + 8𝑥(𝑎 + 𝑏)

4 − 𝑘 ^% (𝑎 + 𝑏)(𝑥 ^" − 6𝑥 ^% + 8𝑥)

(2 + 𝑘)(2 − 𝑘) 𝑥(𝑎 + 𝑏)(𝑥 ^% − 6𝑥 + 8)

(2 + 𝑥 + 7)E2 − (𝑥 + 7)F 𝑥(𝑎 + 𝑏)(𝑥 − 2)(𝑥 − 4) (9 + 𝑥)(2 − 𝑥 − 7)

(9 + 𝑥)(−5 − 𝑥)

−(9 + 𝑥)(5 + 𝑥)

5) −8𝑥 ^" + 10𝑥 ^% + 12𝑥 6) 25𝑥 ^% (𝑎 − 1) ^" − 5𝑥(𝑎 − 1) ^" − 2(𝑎 − 1) ^"

−2𝑥(4𝑥 ^% − 5𝑥 − 6) (𝑎 − 1) ^" (25𝑥 ^% − 5𝑥 − 2)

−2𝑥(4𝑥 + 3)(𝑥 − 2) (𝑎 − 1) ^" (5𝑥 − 2)(5𝑥 + 1)

7) ³

345 − ^"

367 = ^"9

3 ^: 4;345

3

345 − ^"

367 = _(345)(367) ^"9 LCD (𝑥 − 5)(𝑥 + 1) G _𝑥−5 ^𝑥 − _𝑥+1 ³ = _{𝑥 2 −4𝑥−5} ³⁰ H (𝑥 − 5)(𝑥 + 1)

𝑥(𝑥 + 1) − 3(𝑥 − 5) = 30 𝑥 ^% + 𝑥 − 3𝑥 + 15 − 30 = 0 𝑥 ^% − 2𝑥 − 15 = 0

(𝑥 − 5)(𝑥 + 3) = 0 𝑥 = 5 𝑟𝑒𝑗𝑒𝑐𝑡 𝑥 = −3

8) 6𝑥 ^% (3𝑥 − 1) − 𝑥(3𝑥 − 1) = 2(3𝑥 − 1) 11) −𝑥 ^% + 8𝑥 − 17 = 0 6𝑥 ^% (3𝑥 − 1) − 𝑥(3𝑥 − 1) − 2(3𝑥 − 1) = 0 −𝑥 ^% + 8𝑥 = 17

(3𝑥 − 1)(6𝑥 ^% − 𝑥 − 2) = 0 −(𝑥 ^% − 8𝑥 + 16) = 17 − 16 (3𝑥 − 1)(3𝑥 − 2)(2𝑥 + 1) = 0 −(𝑥 − 4) ^% = 1

𝑥 = ⁷ _" 𝑥 = ^% _" 𝑥 = − ⁷ _% (𝑥 − 4) ^% = −1 N(𝑥 − 4) ^% = ±√−1

𝑥 − 4 = ±√𝑖 ^%

9) 𝑥 = ^%±>√@ _% 𝑥 = 4 ± 𝑖

2𝑥 = 2 ± 𝑖√6 2𝑥 − 2 = ±𝑖√6

(2𝑥 − 2) ^% = E±𝑖√6F ^% 12) 2𝑥 ^% − 2√2𝑥 = −1

4𝑥 ^% − 8𝑥 + 4 = 6𝑖 ^% 𝑖 ^% = −1 2E𝑥 ^% − 𝑥√2F = −1

4𝑥 ^% − 8𝑥 + 4 = −6 2 P𝑥 ^% − 𝑥√2 + ⁷ _% Q = −1 + 2 P ⁷ _% Q

4𝑥 ^% − 8𝑥 + 10 = 0 2 P𝑥 − ^√% _% Q ^% = 0

RP𝑥 − ^√% _% Q ^% = 0 𝑥 = ^√%

% exact value or 0.707 approx.

10) 3𝑥 ^% + 18𝑥 − 24 = 0 13) 6𝑥(𝑥 + 1) = (𝑥 + 3) ^%

3𝑥 ^% + 18𝑥 = 24 6𝑥 ^% + 6𝑥 = 𝑥 ^% + 6𝑥 + 9

3(𝑥 ^% + 6𝑥 + 9) = 24 + 3(9) 5𝑥 ^% − 9 = 0

3(𝑥 + 3) ^% = 51 𝑎 = 5, 𝑏 = 0, 𝑐 = −9

(𝑥 + 3) ^% = ⁵⁷ _"

N(𝑥 + 3) ^% = ±√17 𝑥 + 3 = ±√17

𝑥 = −3 ± √17 exact value

or 𝑥 = 1.123 𝑥 = −7.123 approximate value

14) Let the width of the rectangle = w Let the length of the rectangle = 2w - 3 The area of a rectangle is length x width. The area is 20 cm ² .

(2𝑤 − 3)(𝑤) = 20 2𝑤 ^% − 3𝑤 − 20 = 0 (𝑤 − 4)(2𝑤 + 5) = 0 𝑤 = 4 𝑤 = −2.5 𝑟𝑒𝑗𝑒𝑐𝑡

The width of the rectangle is 4 cm and the length of the rectangle is 5 cm.

15) √𝑥 + 3 = 2𝑥 − 1 E√𝑥 + 3F ^% = (2𝑥 − 1) ^% 𝑥 + 3 = 4𝑥 ^% − 4𝑥 + 1 4𝑥 ^% − 5𝑥 − 2 = 0 𝑥 = 4(45)±N(45) ^: 4;(;)(4%)

%(;)

𝑥 = 5±√%56"%

*

𝑥 = ^5±√5U _*

𝑥 = ^56√5U _* 𝑥 = ^54√5U _* 𝑟𝑒𝑗𝑒𝑐𝑡 (𝐸𝑥𝑎𝑐𝑡 𝑉𝑎𝑙𝑢𝑒)

𝑥 = 1.569 𝑥 = −0.319 𝑟𝑒𝑗𝑒𝑐𝑡 (𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑉𝑎𝑙𝑢𝑒)

16) ₃ ⁷ − _34; ^% = 2 LCD is 𝑥(𝑥 − 4) 𝑥(𝑥 − 4) G ⁷ ₃ − _34; ^% = 2H

𝑥 − 4 − 2𝑥 = 2𝑥(𝑥 − 4)

−𝑥 − 4 = 2𝑥 ^% − 8𝑥 2𝑥 ^% − 7𝑥 + 4 = 0 𝑥 = 4(4U)±N(4U) ^: 4;(%)(;)

%(%)

𝑥 = U±√;^4"%

;

𝑥 = ^U±√7U _;

𝑥 = 2.781 𝑥 = 0.719

17) 𝑥 ^% + 𝑘𝑥 + 3𝑘 − 5 = 0 𝑎 = 1, 𝑏 = 𝑘, 𝑐 = 3𝑘 − 5 𝑏 ^% − 4𝑎𝑐 = 0

𝑘 ^% − 4(1)(3𝑘 − 5) = 0

𝑘 ^% − 12𝑘 + 20 = 0

(𝑘 − 10)(𝑘 − 2) = 0

𝑘 = 10 𝑘 = 2

18) 𝑥 ^% − 𝑘𝑥 + 9 = 0 𝑎 = 1, 𝑏 = −𝑘, 𝑐 = 9 𝑏 ^% − 4𝑎𝑐 > 0

(−𝑘) ^% − 4(1)(9) > 0 𝑘 ^% − 36 > 0

(𝑘 + 6)(𝑘 − 6) > 0

𝑘 < −6 − 6 < 𝑘 < 6 𝑘 > 6 𝑘 < −6 𝑜𝑟 𝑘 > 6

19) −2𝑥 ^% + 12𝑥 − 10 = 0 𝑥 = 1 𝑥 = 5

20) 0.25𝑥 ^% + 1.5𝑥 − 1.75 = 0 𝑥 = −7 𝑥 = 1

21)

(2𝑥 + 60)(2𝑥 + 20) = (20)(60) + 516 4𝑥 ^% + 40𝑥 + 120𝑥 + 1200 = 1200 + 516 4𝑥 ^% + 160𝑥 + 1200 − 1200 − 516 = 0 4𝑥 ^% + 160𝑥 − 516 = 0

4(𝑥 ^% + 40𝑥 − 129) = 0 4(𝑥 + 43)(𝑥 − 3) = 0 𝑥 = −43 𝑟𝑒𝑗𝑒𝑐𝑡 𝑥 = 3 The width of the frame is 3 cm.

22) 𝑆. 𝐴. = 2(𝑙𝑤 + 𝑙ℎ + 𝑤ℎ) 350 = 2(𝑥 ^% + 9𝑥 + 9𝑥) 350 = 2𝑥 ^% + 36𝑥 2𝑥 ^% + 36𝑥 − 350 = 0 2(𝑥 ^% + 18𝑥 − 175) = 0 2(𝑥 − 7)(𝑥 + 25) = 0 𝑥 = 7 𝑥 = −25 𝑟𝑒𝑗𝑒𝑐𝑡

The length of the base of the box is 7 cm.

23)

slower 300 km x 300/x

faster 300 km x+10 300/x+10

Longer time - shorter time = difference in time

"99

3 − ₃₆₇₉ ^"99 = 1 LCD is x(x+10) 𝑥(𝑥 + 10) G ^"99

3 − ^"99

3679 = 1H

300(𝑥 + 10) − 300𝑥 = 𝑥(𝑥 + 10) 300𝑥 + 3000 − 300𝑥 = 𝑥 ^% + 10𝑥 𝑥 ^% + 10𝑥 − 3000 = 0

(𝑥 + 60)(𝑥 − 50) = 0 𝑥 = −60 𝑟𝑒𝑗𝑒𝑐𝑡 𝑥 = 50

Pedro's original speed was 50 km/h.

60 cm

20 cm

x

x x x

2x + 60

2x + 20

9 cm

x x

d r t

24) Let one of the pipe's rate of work be equal to ⁷ ₃ Let the other pipe's rate of work be equal to _36" ⁷ Let their combined rate of work be equal to ⁷ _%

7

3 + _36" ⁷ = ⁷ _% LCD is 2𝑥(𝑥 + 3)

2𝑥(𝑥 + 3) G ₃ ⁷ + _36" ⁷ = ⁷ _% H 2(𝑥 + 3) + 2𝑥 = 𝑥(𝑥 + 3) 2𝑥 + 6 + 2𝑥 = 𝑥 ^% + 3𝑥 𝑥 ^% + 3𝑥 − 2𝑥 − 2𝑥 − 6 = 0 𝑥 ^% − 𝑥 − 6 = 0

(𝑥 − 3)(𝑥 + 2) = 0 𝑥 = 3 𝑥 = −2 𝑟𝑒𝑗𝑒𝑐𝑡

The slower pipe takes 6 hours to fill the reservoir by itself.

25) 𝜋𝑟 ^% _d>e − 𝜋𝑟 ^% _fghii = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐹𝑙𝑜𝑤𝑒𝑟 𝐵𝑒𝑑

𝜋(8) ^% − 𝜋𝑥 ^% = 36 64𝜋 − 𝜋𝑥 ^% = 36 64𝜋 − 36 = 𝜋𝑥 ^% 64 − ^"@ _n = 𝑥 ^%

±R64 − ^"@ _n = √𝑥 ^%

7.249 = 𝑥 − 7.249 = 𝑥 𝑟𝑒𝑗𝑒𝑐𝑡 8 − 7.249 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑓𝑙𝑜𝑤𝑒𝑟 𝑏𝑒𝑑

0.751 𝑚𝑒𝑡𝑟𝑒𝑠 𝑖𝑠 𝑡ℎ𝑒 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤𝑒𝑟 𝑏𝑒𝑑.

8 m

x