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# Solving Quadratic Equations by Quadratic Formula Worksheet - Page 3

Solving Quadratic Equations by Quadratic Formula Worksheet
• Page 3
21.
A tennis player hits a ball when it is 8 feet off the ground. The ball is hit with an upward velocity of 8 feet per second. After the ball is hit, its height $h$ (in feet) is modeled by $h$ = -16$t$2 + 8$t$ + 8, where $t$ is the time in seconds. How long will it take the ball to reach the ground?
 a. 3 b. 4 c. 1 d. 2

#### Solution:

h = -16t2 + 8t + 8
[Original equation.]

0 = -16t2 + 8t + 8
[Substitute 0 for h, since at ground level the height h = 0.]

t = {-8 ± √[(8)2 - 4(-16)(8)]}/[2(-16)]
[Substitute the values in the quadratic formula: a = -16, b = 8 and c = 8.]

= -8±√(64+512) / -32
[Simplify.]

= -8±√576 / -32
[Simplify inside the radical.]

= -8±24 / -32
[Simplify the radical.]

= -32 / -32 = 1
[Since t represents time, discard negative value.]

Correct answer : (3)
22.
Chris drops a ball from a height of 100 feet above the ground. Calculate the time taken by the ball to hit the ground, if its height is given by the equation $h$ = -16$t$2 + 100, where $t$ is the time in seconds.
 a. 3 b. 2.25 c. 2.5 d. 2.75

#### Solution:

h = -16t2 + 100.
[Original equation.]

0 = -16t2 + 100
[Substituting 0 for h, since at ground level h = 0.]

t = {-0 ± √[(-0)2 - 4(-16)(100)]}/[2(-16)]
[Substitute the values in the quadratic formula: a = -16, b = 0 and c = 100.]

= 0±√(0+6400) / -32
[Simplify.]

= 0±√6400 / -32
[Simplify inside the grouping.]

= 0±80 / -32
[Simplify the radical.]

= -80 / -32 = 2.5
[Since t represents time, rounding it to a positive value.]

So, the ball takes 2.5 sec to reach the ground.

Correct answer : (3)
23.
Andy throws a pencil from a building with an initial downward velocity of - 8 feet per second. How long will the pencil take to reach the ground, if the height of the pencil from the ground is modeled by the equation $h$ = - 16$t$2 - 8$t$ + 48, where $t$ is the time in seconds?
 a. 1.5 b. 2 c. 1.75 d. 1.25

#### Solution:

h = - 16t2 - 8t + 48
[Original equation.]

0 = - 16t2 - 8t + 48
[Substitute values and write in the standard form.]

t = {- (- 8) ± √[(- 8)2 - 4(- 16)(48)]}/2(- 16)
[Substitute the values in the quadratic formula.]

= 8±√(64+3072) / -32
[Simplify.]

= 8±√3136 / -32
[Simplify the radical.]

= 8±56 / -32
[Simplify.]

= -48 / -32 = 1.5
[Evaluating the radical and rounding the solution to a positive value as t represents time.]

Correct answer : (1)
24.
What are the values of $a$, $b$ and $c$ in the equation 4$f$2 + 3$f$ - 38 = 0, which is in the standard form?
 a. $a$ = -4, $b$ = 3 and $c$ = 38 b. $a$ = 4, $b$ = 3 and $c$ = -38 c. $a$ = 4, $b$ = -3 and $c$ = 38 d. None of the above

#### Solution:

The standard equation is ax2 + bx + c = 0 when a ≠ 0.

4f2 + 3f - 38 = 0
[Original equation.]

a = 4, b = 3 and c = -38
[Compare the original equation with the standard equation.]

Correct answer : (2)
25.
Write the equation $\frac{1}{3}$$g$2 - 3 = - $\frac{14}{15}$$g$ in the standard form.
 a. 5$g$2 + 14$g$ - 45 = 0 b. -5$g$2 - 14$g$ - 45 = 0 c. 5$g$2 - 14$g$ - 45 = 0 d. None of the above

#### Solution:

13g2 - 3 = - 1415g
[Original equation.]

13g2 - 3 + 1415g = 0
[Add 14 / 15g to each side.]

5g2 - 45 + 14g = 0
[Multiply the equation by LCM, 15.]

5g2 + 14g - 45 = 0
[Rewrite the equation in the standard form.]

Correct answer : (1)

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