﻿ Quadratic Formula Worksheet | Problems & Solutions

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1.
Solve the equation $\frac{1}{4}$$k$2 - 4$k$ + 12 = 0 by using the quadratic formula.
 a. $k$ = - 4, 12 b. $k$ = - 4, - 12 c. $k$ = 4, 12 d. $k$ = 4, - 12

#### Solution:

1 / 4k2 - 4k + 12 = 0
[Original equation.]

k2 - 16k + 48 = 0
[Multiply the equation by 4.]

k = {-(-16)±[(-16)2-4(1)(48)]}2(1)
[Substitute a = 1, b = - 16 and c = 48 in the quadratic formula.]

k = [16±(64)]2

k = (16+8)2, (16-8)2
[Write ± as two separate equations.]

k = 12, 4
[Simplify.]

2.
Solve the system:
$x$2 + $y$2 = 17
$y$ - $x$ = 4

#### Solution:

x2 + y2 = 17. . . (1)
y - x = 4. . . (2)
[Original system of equations.]

y = x + 4
[Rearrange eq (2).]

y2 = x2 + 8x + 16 ----(3)
[Use (a + b)2 = a2 + 2ab + b2.]

x2 + (x2 + 8x + 16) = 17
[Substitute y2 from eq (3) in eq (1).]

2x2 + 8x - 1 = 0
[Simplify.]

x = - (8) ±(8)2+4(2)(1)2(2)
[Use the quadratic formula - b ±b2-4ac2a.]

= - 8±724
[Simplify.]

= - 2 ± 182
[Simplify.]

y = x + 4

y = - 2 + 182 + 4
[Substitute x = - 2 + 182.]

y = 2 + 182
[Simplify.]

y = x + 4

y = - 2 - 182 + 4
[Substitute x = - 2 - 182.]

y = 2 - 182
[Simplify.]

So, the solutions are (- 2 + 182, 2 + 182) and (- 2 - 182, 2 - 182).

3.
Which of the following are the solutions of the quadratic equation $\mathrm{ax}$2 + $\mathrm{bx}$ + $c$ = 0, when $a$ ≠ 0 and $b$2 - 4$\mathrm{ac}$ ≥ 0?
 a. $x$ = b. $x$ = c. $x$ = - $\frac{b}{a}$ and $x$ = $\frac{c}{a}$ d. $x$ = - $\frac{a}{b}$ and $x$ = $\frac{c}{b}$

#### Solution:

The solutions of the quadratic equation ax2 + bx + c = 0 are x = -b±b2-4ac2a when a ≠ 0 and b2 - 4ac ≥ 0.

4.
Write the equation - $d$2 + 3 = 5$d$ - 5$d$2 in the standard form.
 a. 4$d$2 + 5$d$ - 3 = 0 b. 4$d$2 - 5$d$ + 3 = 0 c. 4$d$2 - 5$d$ - 3 = 0 d. 4$d$2 + 5$d$ + 3 = 0

#### Solution:

- d2 + 3 = 5d - 5d2
[Original equation.]

4d2 + 3 = 5d

4d2 + 3 - 5d = 0
[Subtract 5d from each side.]

4d2 - 5d + 3 = 0
[Rewrite the equation in the standard form.]

5.
What are the values of $a$, $b$ and $c$ in the quadratic formula used to solve the equation 5$e$2 - 4 + $e$ = - $e$2 + 5$e$?
 a. $a$ = 6, $b$ = - 4 and $c$ = 4 b. $a$ = 6, $b$ = 4 and $c$ = - 4 c. $a$ = 6, $b$ = - 4 and $c$ = - 4 d. $a$ = - 6, $b$ = - 4 and $c$ = - 4

#### Solution:

5e2 - 4 + e = - e2 + 5e
[Original equation.]

6e2 - 4 + e = 5e

6e2 - 4 - 4e = 0
[Subtract 5e from each side.]

6e2 - 4e - 4 = 0
[Rewrite the equation in the standard form.]

In the quadratic formula, a is the coefficient of the term of degree 2, b is the coefficient of the term of degree 1 and c is the constant term.

So, a = 6, b = - 4 and c = - 4.

6.
What are the values of $a$, $b$ and $c$ in the equation 4$f$ 2 - 2$f$ + 33 = 0, which is in the standard form?
 a. $a$ = 4, $b$ = - 2 and $c$ = 33 b. $a$ = 4, $b$ = 2 and $c$ = 33 c. $a$ = 4, $b$ = 2 and $c$ = - 33 d. $a$ = - 4, $b$ = - 2 and $c$ = - 33

#### Solution:

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

4f 2 - 2f + 33 = 0
[Original equation.]

a = 4, b = - 2 and c = 33
[Compare the original equation with the standard equation.]

7.
Write the equation $\frac{1}{3}$$g$2 - 2 = - $\frac{11}{12}$$g$, in the standard form.
 a. $\frac{1}{3}$$g$2 - $\frac{11}{12}$$g$ + 2 = 0 b. $\frac{1}{3}$$g$2 + $\frac{11}{12}$$g$ - 2 = 0 c. $\frac{1}{3}$$g$2 - $\frac{11}{12}$$g$ - 2 = 0 d. $\frac{1}{3}$$g$2 + $\frac{11}{12}$$g$ + 2 = 0

#### Solution:

13g2 - 2 = - 1112g
[Original equation.]

13g2 - 2 + 1112g = 0
[Add 11 / 12g to both sides.]

13g2 + 1112g - 2 = 0
[Rewrite the equation in the standard form.]

8.
Find the value of $b$2 - 4$a$$c$, for the equation -8$x$2 - 20$x$ - 12 = 0.
 a. 784 b. - 784 c. 404 d. 16

#### Solution:

-8x2 - 20x - 12 = 0
[Original equation.]

8x2 + 20x + 12 = 0
[Rewrite the equation in the standard form.]

a = 8, b = 20 and c = 12.
[Compare the equation with standard equation ax2 + bx + c = 0.]

b2 - 4ac = 202 - 4(8)(12)
[Replace a with 8, b with 20 and c with 12.]

= 16
[Simplify.]

9.
Find the solutions of the quadratic equation $p$2 + 7$p$ + 12 = 0.
 a. - 3, - 4 b. 3, 4 c. 7, 12 d. - 19

#### Solution:

p2 + 7p + 12 = 0
[Original equation.]

p = {-7±[(7)2-4(1)(12)]}2(1)
[Substitute a = 1, b = 7 and c = 12 in the quadratic formula.]

p = [-7±(49-48)]2
[Simplify.]

p = (-7±1)2
[Simplify.]

p = (-7+1)2, (-7-1)2
[Simplify.]

p = - 3, - 4

10.
Jeff throws a pen from the top of a 124 feet tall building with an initial downward velocity of - 30 feet per second. How long will the pen take to reach the ground?
 a. 2 b. 30 c. 124 d. - 3.785

#### 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)