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Quadratic Formula Worksheet

Quadratic Formula Worksheet
  • Page 1
 1.  
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±(64)]2
[Simplify inside the radical.]

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

k = 12, 4
[Simplify.]


Correct answer : (3)
 2.  
Solve the system:
x2 + y2 = 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).


Correct answer : (0)
 3.  
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- 4ac2a
b.
x = b±b2- 4ac2
c.
x = - ba and x = ca
d.
x = - ab and x = cb


Solution:

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


Correct answer : (1)
 4.  
Write the equation - d2 + 3 = 5d - 5d2 in the standard form.
a.
4d2 + 5d - 3 = 0
b.
4d2 - 5d + 3 = 0
c.
4d2 - 5d - 3 = 0
d.
4d2 + 5d + 3 = 0


Solution:

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

4d2 + 3 = 5d
[Add 5d2 to both sides.]

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

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


Correct answer : (2)
 5.  
What are the values of a, b and c in the quadratic formula used to solve the equation 5e2 - 4 + e = - e2 + 5e?
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
[Add e2 to both sides.]

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.


Correct answer : (3)
 6.  
What are the values of a, b and c in the equation 4f 2 - 2f + 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.]


Correct answer : (1)
 7.  
Write the equation 1 3g2 - 2 = - 11 12g, in the standard form.
a.
1 3g2 - 11 12g + 2 = 0
b.
1 3g2 + 11 12g - 2 = 0
c.
1 3g2 - 11 12g - 2 = 0
d.
1 3g2 + 11 12g + 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.]


Correct answer : (2)
 8.  
Find the value of b2 - 4ac, for the equation -8x2 - 20x - 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.]


Correct answer : (4)
 9.  
Find the solutions of the quadratic equation p2 + 7p + 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


Correct answer : (1)
 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)
[Simplify the radical.]

= - 3.875, 2
[Simplify.]

The ball reaches the ground after 2 seconds.
[Consider positive value as t represents time.]


Correct answer : (1)

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