﻿ Solving Quadratic Equations by Quadratic Formula Worksheet | Problems & Solutions

• Page 1
1.
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 ± √(b2 - 4ac))/(2a) b. $x$ = (-b ± √(b2 - 4ac))/2 c. $x$ = (b ± √(b2 - 4ac))/(2a) d. None of the above

#### Solution:

The solutions of the quadratic equation ax2 + bx + c =0 are x = (-b ± √(b2 - 4ac))/(2a) when a ≠ 0 and b2 - 4ac ≥ 0.

2.
Write the equation -$d$2 + 3 = 7$d$ - 6$d$2 in the standard form.
 a. 5$d$2 + 7$d$ + 3 = 0 b. 5$d$2 - 7$d$ - 3 = 0 c. 5$d$2 - 7$d$ + 3 = 0 d. 5$d$2 + 7$d$ - 3 = 0

#### Solution:

-d2 + 3 = 7d - 6d2
[Original equation.]

5d2 + 3 = 7d

5d2 + 3 - 7d = 0
[Subtract 7d from each side.]

5d2 - 7d + 3 = 0
[Rewrite the equation in the standard form.]

3.
What are the values of $a$, $b$ and $c$ in the quadratic formula used to solve the equation 8$e$2 - 4 + $e$ = - $e$2 + 6$e$?
 a. $a$ = -9, $b$ = -5 and $c$ = -4 b. $a$ = 9, $b$ = -5 and $c$ = 4 c. $a$ = 9, $b$ = -5 and $c$ = -4 d. None of the above

#### Solution:

8e2 - 4 + e = - e2 + 6e
[Original equation.]

9e2 - 4 + e = 6e

9e2 - 4 - 5e = 0
[Subtract 6e from each side.]

9e2 - 5e - 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 = 9, b = -5 and c = -4

4.
What are the values of $a$, $b$ and $c$ in the equation 2$f$2 - 7$f$ + 31 = 0, which is in the standard form?
 a. $a$ = 2, $b$ = -7 and $c$ = 31 b. $a$ = 2, $b$ = 7 and $c$ = 31 c. $a$ = -2, $b$ = -7 and $c$ = -31 d. None of the above

#### Solution:

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

2f2 - 7f + 31 = 0
[Original equation.]

a = 2, b = -7 and c = 31
[Compare the original equation with the standard equation.]

5.
Write the equation $\frac{1}{2}$$g$2 - 1 = - $\frac{7}{8}$$g$ in the standard form.
 a. 4$g$2 - 7$g$ - 8 = 0 b. 4$g$2 + 7$g$ - 8 = 0 c. 4$g$2 + 7$g$ + 8 = 0 d. 4$g$2 - 7$g$ + 8 = 0

#### Solution:

12g2 - 1 = - 78g
[Original equation.]

12g2 - 1 + 78g = 0
[Add 7 / 8g on each side.]

4g2 - 8 + 7g = 0
[Multiply the equation by LCM, 8.]

4g2 + 7g - 8 = 0
[Rewrite the equation in the standard form.]

6.
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 x j 2 - 1 = 12 x j + 1
[Original equation.]

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

14 x j 2 - 2 - 12 x 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 x 4 = 28 or 7 x -2 = -14
[Substitute j and simplify.]

7.
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 ±642

k = 16 + 82, 16 - 82
[Write ± as two separate equations.]

k = 12, 4
[Simplify.]

8.
Find the solutions of the quadratic equation $p$2 + 8$p$ + 15 = 0.
 a. 3, 5 b. - 3, 5 c. 3, - 5 d. - 3, - 5

#### Solution:

p2 + 8p + 15 = 0
[Original equation.]

p = - 8 ±82 - 4(1)(15)2(1)
[Substitute a = 1, b = 8 and c = 15 in the quadratic formula.]

p = - 8 ±(64 - 60)2
[Simplify.]

p = - 8 ±42
[Simplify.]

p = (- 8 + 2)2, (- 8 - 2)2
[Simplify.]

p = - 3, - 5

9.
Write the equation ($\frac{4}{3}$× $h$2) + 6 = $h$2 + ($\frac{16}{15}$× $h$) in the standard form and identify the values of $a$, $b$ and $c$.
 a. $a$ = 5, $b$ = 16 and $c$ = 90 b. $a$ = 5, $b$ = -16 and $c$ = -90 c. $a$ = 5, $b$ = -16 and $c$ = 90 d. None of the above

#### Solution:

(43x h2) + 6 = h2 + (1615x h)
[Original equation.]

(13x h2) + 6 = (1615x h)
[Subtract h2 from each side.]

(13x h2) + 6 - (1615x h) = 0
[Subtract (16 / 15x h) from each side.]

5h2 + 90 - 16h = 0
[Multiply the equation with the LCM, 15.]

5h2 - 16h + 90 = 0
[Rewrite the equation in the standard form.]

a = 5, b = -16 and c = 90.
[Compare the equation with the standard form, ax2 + bx + c = 0.]

10.
Find the value of $b$2 - 4$\mathrm{ac}$ for the equation -6$x$2 - 16$x$ - 8 = 0.
 a. -64 b. 64 c. 448 d. None of the above

#### Solution:

-6x2 - 16x - 8 = 0
[Original equation.]

6x2 + 16x + 8 = 0
[Rewrite the equation in the standard form.]

a = 6, b = 16 and c = 8.
[Compare the equation with standard equation ax2 + bx + c = 0.]

b2 - 4ac = 162 - 4(6)(8)
[Substitute 6 for a, 16 for b and 8 for c.]

= 64
[Simplify.]