﻿ Solving Quadratic Equations by Quadratic Formula Worksheet - Page 2 | Problems & Solutions

• Page 2
11.
Compare $\frac{3}{4}$$n$2 - 4 = $\frac{1}{2}$$n$ - 1 with the standard form and find the value of $b$2 - 4$\mathrm{ac}$.
 a. 148 b. -148 c. -140 d. 140

#### Solution:

34n2 - 4 = 12n - 1
[Original equation.]

34n2 - 4 - 12n = -1
[Subtract 1 / 2n from each side.]

34n2 - 12n - 3 = 0

3n2 - 2n - 12 = 0
[Multiply the equation by 4.]

a = 3, b = -2 and c = -12
[Compare the equation with the standard form ax2 + bx + c and find the values.]

b2 - 4ac = (-2)2 - 4(3)(-12)
[Substitute the values.]

= 148
[Simplify.]

12.
Which of the following quadratic equations has the solutions [11 ± √(121 + 40)] / 10?
 a. 5$x$2 - 11$x$ - 2 = 0 b. 5$x$2 + 11$x$ - 2 = 0 c. 5$x$2 + 11$x$ + 2 = 0 d. None of the above

#### Solution:

Compare [11 ± √(121 + 40)] / 10 with [-b ± √(b2 - 4ac)]/(2a)

2a = 10. So, a = 5.

-b = 11. So, b = -11.

-4ac = 40. So, c = -2.
[Substitute a = 5 and simplify.]

ax2 + bx + c = 5x2 + (-11)x + (-2)
[Substitute the values in the standard form.]

= 5x2 - 11x - 2
[Simplify.]

So, the quadratic equation is 5x2 - 11x - 2 = 0.

13.
Jake throws a pen from the top of a building with an initial downward velocity of -30 feet per second. How long will the pen take to reach the ground, if the height of the pen from the ground is modeled by the equation $h$ = -16$t$2 - 30$t$ + 124, where $h$ is the height of the pen and $t$ is the time in seconds?
 a. 124 b. 2 c. -3 d. 30

#### Solution:

h = -16t2 - 30t + 124
[Original equation.]

0 = -16t2 - 30t + 124
[Height = 0, when the pen is on the ground.]

Compare the original equation with the standard form to get the values of a, b and c.

t = [-(-30) ± √[(-30)2 - 4(-16)(124)]] / [2(-16)]
[Substitute the values in the quadratic formula.]

t = [30 ± √(900 + 7936)]/(-32)
[Evaluate power and multiply.]

= [30 ± 94]/(-32)

= - 3.875, 2
[Simplify.]

The ball reaches the ground after 2 seconds.
[Consider positive value as t represents time.]

14.
Find the $x$-intercepts of the graph of $y$ = -$x$2 - 3$x$ + 18.
 a. 3, -6 b. 3, 6 c. -3, 6 d. -3, -6

#### Solution:

The x-intercepts occur when y = 0.

y = -x2 - 3x + 18
[Original equation.]

0 = x2 + 3x - 18
[Substitute y = 0 and write the equation in the standard form.]

x = {-3 ± √[(3)2 - 4(1)(-18)]} / 2(1)
[Substitute a = 1, b = 3 and c = -18 in the quadratic formula.]

= -3±√(9+72) / 2
[Simplify the squares.]

= -3±√81 / 2

= -3±9 / 2

= -3+9 / 2 , -3-9 / 2
[Write ± as two separate terms.]

= 3, -6
[Simplify.]

15.
Olga dives into a pool from the diving board, which was 4 feet high from the water. She dives with an initial downward velocity of -12 feet per second. If the equation to model the height of the dive is $h$ = -16$t$2 +(-12)$t$ + 4, then find the time (in seconds) Olga takes to reach the water level.
 a. 0.25 b. 2.25 c. 1.25 d. None of the above

#### Solution:

h = -16t2 + (-12)t + 4
[Original equation.]

0 = -16t2 + (-12t) + 4
[Substitute 0 for h, since at water level the height is zero.]

t = {-(-12) ± √[(-12)2 - 4(-16)(4)]}/[2(-16)]
[Substitute a = -16, b = -12 and c = 4 in the quadratic formula.]

t = 12±√(144+256) / -32
[Simplify.]

t = 12±√400 / -32

t = 12±20 / -32 = -1, 0.25

t = 0.25
[Since t represents time, consider the positive integer.]

16.
Edward stands on a bridge 73.5 feet above the ground holding an apple. He throws it with an initial downward velocity of -25 feet per second. How long will it take for the apple to reach the ground, if the vertical motion is given by the equation $h$ = -16$t$2 + $v$$t$ + $s$?
 a. 1.5 seconds b. 3.06 seconds c. 2 seconds d. 2.5 seconds

#### Solution:

h = -16t2 + vt + s
[Original equation.]

0 = -16t2 + vt + s
[h = 0 for ground level.]

0 = -16t2 - 25t + 73.5
[Replace v with -25 and s = 73.5.]

t = [-(-25) ±[(-25)² - 4(-16)(73.5)]]2(-16)
[Substitute the values of a = -16, b = -25 and c = 73.5 in the quadratic formula.]

= [25 ±(625 + 4704)]-32
[Evaluate the power and multiply.]

= 25 ±5329-32

= 25 ± 73-32
[Find the square root.]

t = -3.0625 or 1.5
[Simplify.]

The apple will reach the ground about 1.5 seconds after it was thrown.

17.
Find the $x$-intercepts of the graph of $y$ = $x$2 + 5$x$ - 24.
 a. -8, -3 b. 8, -3 c. -8, 3 d. None of the above

#### Solution:

y = x2 + 5x - 24
[Original equation.]

0 = x2 + 5x - 24
[Substitute y = 0 to find the x-intercepts.]

= -5±√(25+96) / 2
[Substitute the values of a, b and c in the quadratic formula.]

= -5±√121 / 2

= -5±11 / 2
[Evaluate the power.]

x = -8 or 3
[Simplify.]

So, the x-intercepts are -8 and 3.

18.
Sunny jumped from a bungee tower, which was 784 feet high. Find the time taken by him to reach the ground, if the equation that models his height is $h$ = -16$t$2 + 784, where $t$ is the time in seconds.
 a. 6.75 b. 7.25 c. 7 d. 7.5

#### Solution:

h = -16t2 + 784
[Original equation.]

0 = -16t2 + 784
[Substitute 0 for h, since at ground level the height is zero.]

t = {-(-0) ± √[(-0)2 - 4(-16)(784)]}/[2(-16)]
[Substitute the values in the quadratic formula: a = -16, b = 0 and c = 784.]

= 0±√(0+50176)] / -32
[Simplify.]

= 0±√50176 / -32

= 0±224 / -32
[Simplify.]

= -224 / -32 = 7
[Since t represents time, rounding t to a positive number.]

19.
A stone is dropped from a height of 9 feet above the ground. The height of the stone is modeled by the equation $h$ = -16$t$2 + 9, where $t$ is the time in seconds. Find the time taken for the stone to hit the ground.
 a. 0.5 b. 1 c. 1.25 d. 0.75

#### Solution:

h = -16t2 + 9
[Original equation.]

0 = -16t2 + 9
[Substitute 0 for h, since at ground level the height is zero.]

t = {-0 ± √[02 - 4(-16)(9)]}/(2(-16))
[Substitute the values in the quadratic formula: a = -16, b = 0 and c = 9.]

= 0±√(0+576) / -32
[Simplify.]

0±√576 / -32

= 0±24 / -32

t = -24 / -32 = 0.75
[Since t represents time, rounding it to a positive number.]

20.
The cost price of cotton purchased by a textile company is given by the function $p$ = $s$2 + 2$s$ - 1520, where $s$ is the number of bales purchased. How many bales of cotton are to be purchased for the cost price to be minimum?
 a. 40 b. 38 c. 42 d. 44

#### Solution:

p = s2 + 2s - 1520
[Original equation.]

0 = s2 + 2s - 1520
[Substitute 0 for p, since for the minimum price, the equation should be equal to zero.]

s = {-(2) ± √[22 - 4(1)(-1520)]}/2(1)
[Substitute the values in the quadratic formula: a = 1, b = 2 and c = -1520.]

= -2±√(4+6080) / 2
[Simplify.]

= -2±√6084 / 2