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Solving Quadratic Equations by Quadratic Formula Worksheet

Solving Quadratic Equations by Quadratic Formula Worksheet
  • Page 1
 1.  
Which of the following are the solutions of the quadratic equation ax2 + bx + c = 0, when a ≠ 0 and b2 - 4ac ≥ 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.


Correct answer : (1)
 2.  
Write the equation -d2 + 3 = 7d - 6d2 in the standard form.
a.
5d2 + 7d + 3 = 0
b.
5d2 - 7d - 3 = 0
c.
5d2 - 7d + 3 = 0
d.
5d2 + 7d - 3 = 0


Solution:

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

5d2 + 3 = 7d
[Add 6d2 on each side.]

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

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


Correct answer : (3)
 3.  
What are the values of a, b and c in the quadratic formula used to solve the equation 8e2 - 4 + e = - e2 + 6e?
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
[Add e2 to each side.]

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


Correct answer : (3)
 4.  
What are the values of a, b and c in the equation 2f2 - 7f + 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.]


Correct answer : (1)
 5.  
Write the equation 1 2g2 - 1 = - 7 8g in the standard form.
a.
4g2 - 7g - 8 = 0
b.
4g2 + 7g - 8 = 0
c.
4g2 + 7g + 8 = 0
d.
4g2 - 7g + 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.]


Correct answer : (2)
 6.  
Find the values of 7j in the equation 1 4 × j 2 - 1 = 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.]


Correct answer : (1)
 7.  
Solve the equation 1 4k2 - 4k + 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
[Simplify inside the radical.]

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

k = 12, 4
[Simplify.]


Correct answer : (4)
 8.  
Find the solutions of the quadratic equation p2 + 8p + 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


Correct answer : (4)
 9.  
Write the equation (4 3× h2) + 6 = h2 + (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.]


Correct answer : (3)
 10.  
Find the value of b2 - 4ac for the equation -6x2 - 16x - 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.]


Correct answer : (2)

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