﻿ Quadratic Formula Worksheet - Page 2 | Problems & Solutions

# Quadratic Formula Worksheet - Page 2

• Page 2
11.
Rewrite the equation $\frac{3{h}^{2}}{2}$ + 5 = $h$2 + ($\frac{5h}{4}$), into the standard form and identify the values of $a$, $b$ and $c$.
 a. $a$ = $\frac{1}{2}$, $b$ = $\frac{5}{4}$ and $c$ = 5 b. $a$ = - 1 , $b$ = - 5 and $c$ = - 5 c. $a$ = $\frac{1}{2}$, $b$ = - $\frac{5}{4}$ and $c$ = - 5 d. $a$ = $\frac{1}{2}$, $b$ = - $\frac{5}{4}$ and $c$ = 5

#### Solution:

3h22 + 5 = h2 + (5h4)
[Original equation.]

(h22) + 5 = (5h4)
[Subtract h2 from each side.]

(h22) - (5h4) + 5 = 0
[Rewrite the equation in the standard form.]

a = 1 / 2, b = - 5 / 4 and c = 5
[Compare the equation with the standard form, ax2 + bx + c = 0.]

12.
Find the values of 7$j$, in the equation $\frac{1}{4}$ $j$ 2 - 1 = $\frac{1}{2}$ $j$ + 1.
 a. - 28, 14 b. 28, 14 c. 28, - 14 d. - 28, - 14

#### Solution:

14 j 2 - 1 = 12 j + 1
[Original equation.]

14 j 2 - 1 - 12 j = 1
[Subtract 12 j from each side.]

14 j 2 - 2 - 12 j = 0
[Subtract 1 from each side.]

j2 - 8 - 2j = 0
[Multiply the equation with 4.]

j2 - 2j - 8 = 0
[Rewrite the equation in the standard form as ax2 + bx + c = 0.]

j = 4 or - 2
[Solve for j.]

7j = 7 × 4 = 28 or 7 × - 2 = - 14
[Substitute j and simplify.]

13.
Solve the equation $\frac{1}{3}$$k$2 - 3$k$ + 6 = 0 by using the quadratic formula.
 a. $k$ = 3, 6 b. $k$ = - 3, - 6 c. $k$ = 9, 18 d. $k$ =

#### Solution:

1 / 3k2 - 3k + 6 = 0
[Original equation.]

k2 - 9k + 18 = 0
[Multiply the equation by 3.]

k = -(- 9) ±[(- 9)² - 4(1)(18)]2(1)
[Substitute a = 1, b = - 9 and c = 18 in the quadratic formula.]

k = 9 ±92

k = (9 + 3)2, (9 - 3)2
[Write ± as two separate equations.]

k = 6, 3
[Simplify.]

14.
Compare $\frac{11}{12}$$n$2 - 3 = $\frac{1}{3}$$n$ - 1 with the standard form and find the value of $b$2 - 4$\mathrm{ac}$.
 a. - 1072 b. 1040 c. 1072 d. -1040

#### Solution:

11 / 12n2 - 3 = 1 / 3n - 1
[Original equation.]

11 / 12n2 - 3 - 1 / 3n = - 1
[Subtract 1 / 3n from each side.]

11 / 12n2 - 1 / 3n - 2 = 0

11n2 - 4n - 24 = 0
[Multiply the equation by 12.]

a = 11, b = - 4 and c = - 24
[Compare the equation with the standard form ax2 + bx + c = 0 and find the values.]

b2 - 4ac = (- 4)2 - 4(11)(- 24)
[Substitute the values.]

= 1072
[Simplify.]

15.
Which of the following quadratic equations has the solutions $\frac{\left[11±\sqrt{\left(121+100\right)}\right]}{10}$?
 a. 5$x$2 + 11$x$ - 5 = 0 b. 5$x$2 - 11$x$ - 5 = 0 c. 5$x$2 + 11$x$ + 5 = 0 d. 5$x$2 - 11$x$ + 5 = 0

#### Solution:

Compare [11±(121+100)]10 with [-b±(b2-4ac)](2a)

2a = 10. So, a = 5.

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

- 4ac = 100. So, c = - 5.
[Substitute a = 5 and simplify.]

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

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

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

16.
Find the $x$-intercepts of the graph of $y$ = - $x$2 - 3$x$ + 10.
 a. - 2, - 5 b. 2, 5 c. - 2, 5 d. 2, - 5

#### Solution:

The x-intercepts occur when y = 0.

y = - x2 - 3x + 10
[Original equation.]

0 = x2 + 3x - 10
[Substitute y = 0 and write the equation in the standard form.]

x = {- 3 ±[(3)² - 4(1)(-10)]}2(1)
[Substitute a = 1, b = 2 and c = - 10 in the quadratic formula.]

= - 3±(9+40)2

= - 3±492

= - 3±72

= - 3 + 72 and - 3  -  72
[Write the expression as two separate terms.]

= 2 and - 5
[Simplify.]

So, the x-intercepts of the graph of y = - x2 - 3x + 10 are 2 and - 5.
[Simplify.]

17.
Diane dives into a pool from the diving board, which was 9 feet high from the water. She dives with an initial downward velocity of - 18 feet per second. If the equation to model the height of the dive is $h$ = - 16$t$2 + (- 18)$t$ + 9, then find the time in seconds taken by Diane to reach the water level.
 a. 5.37 b. 0.74 c. 1.37 d. 0.37

#### Solution:

h = - 16t2 + (- 18)t + 9
[Original equation.]

0 = - 16t2 + (- 18t) + 9
[Replace h with 0, as the height is zero at the water level.]

t = {-(-18)±[(-18)2-4(-16)(9)]}[2(-16)]
[Substitute a = - 16, b = - 18 and c = 9 in the quadratic formula.]

t = 18±324+576-32
[Simplify.]

t = 18±900-32

t = 18±30 / -32 = -1.50, 0.37

t = 0.37
[Since t represents time, consider the positive integer.]

18.
Tony 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 + $\mathrm{vt}$ + $s$.
($s$ = 73.5 feet)
 a. 3.06 seconds b. 1.5 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.

19.
Find the $x$-intercepts of the graph of $y$ = $x$2 + 6$x$ - 55.
 a. - 11, - 5 b. - 11, 5 c. 11, - 5 d. 11, 5

#### Solution:

y = x2 + 6x - 55
[Original equation.]

0 = x2 + 6x - 55
[Substitute y = 0 to find the x-intercepts.]

x = (-6)±36+2202
[Substitute the values of a, b and c in the quadratic formula.]

= - 6±2562

= - 6±162

= - 11 and 5
[Separate the terms and simplify.]

So, the x-intercepts are - 11 and 5.

20.
Frank jumped from a bungee tower, which was 729 feet high. Find the time taken by him to reach the ground, if the equation that models his height is $h$ = - 16$t$2 + 729, where $t$ is the time in seconds.
 a. 7.25 b. 6.5 c. 6.75 d. 7

#### Solution:

h = - 16t2 + 729
[Original equation.]

0 = - 16t2 + 729
[Replace h with 0, as the height is zero at the ground level.]

t = {-(-0)±[(-0)2-4(-16)(729)]}[2(-16)]
[Substitute the values in the quadratic formula: a = - 16, b = 0 and c = 729.]

= 0±(0+46656)-32
[Simplify.]

= 0±46656-32