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Solve Quadratic Equations by Finding Square Roots Worksheet

Solve Quadratic Equations by Finding Square Roots Worksheet
  • Page 1
 1.  
Solve: 4w2 - 64 = 0
a.
6, - 6
b.
- 4
c.
4 , - 4
d.
5, - 5


Solution:

4w 2 - 64 = 0
[Original equation.]

4w 2 = 64
[Add 64 to both sides.]

w 2 = 16
[Divide each side by 4.]

w = ± 16
[Evaluate square root on both sides.]

w = ± 4
[Simplify.]


Correct answer : (3)
 2.  
Find the value of x, if 9x2 = 225.
a.
± 6
b.
± 8
c.
± 5
d.
± 7


Solution:

9x2 = 225
[Original equation.]

9x29 = 2259
[Divide each side by 9.]

x2 = 25
[Simplify.]

x = ± 25
[Find square roots.]

x = ± 5
[Evaluate square root.]


Correct answer : (3)
 3.  
Solve 5z2 + 47 = 217 and round the result to its nearest tenth.
a.
8.1
b.
± 8.1
c.
± 5.8
d.
± 7.5


Solution:

5z2 + 47 = 217
[Original equation.]

5z2 = 170
[Subtract 47 from each side.]

z2 = 34
[Divide with 5 on both sides.]

z = ± 5.8
[Take square root on both sides and round to the nearest tenth.]


Correct answer : (3)
 4.  
What integral values of x satisfy the equation x2 = 16?
a.
4, - 4
b.
- 4
c.
16
d.
5, - 5


Solution:

x2 = 16
[Original equation.]

x = ± 16
[Take square root on both sides.]

x = ± 4
[42 = 16 and (- 4)2 = 16.]

The values of x satisfying the equation are 4 and - 4.


Correct answer : (1)
 5.  
Express the solutions of the equation m2 - 15 = 21, as integers.
a.
+ 6 only
b.
+ 6, - 6
c.
+ 36, - 36
d.
- 6 only


Solution:

m2 - 15 = 21
[Original equation.]

m2 = 36
[Add 15 to both sides.]

m = ± 36
[Evaluate square root on both sides.]

m = ± 6
[Simplify.]

The solutions of the equation are + 6 and - 6.


Correct answer : (2)
 6.  
What is the value of x, if 2x2 - 3 = 69?
a.
6
b.
36
c.
± 6
d.
5


Solution:

2x2 - 3 = 69
[Original equation.]

2x2 - 3 + 3 = 69 + 3
[Add 3 to both sides.]

2x2 = 72
[Simplify each side.]

x2 = 36
[Divide each side by 2.]

x = 36
[Find the square roots.]

x = ± 6
[Since 62 = 36 and (-6)2 = 36.]

The value of x, if 2x2 - 3 = 69, is ± 6.


Correct answer : (3)
 7.  
Find the side of a square floor, if its area is 256 square meters.
a.
15 m
b.
14 m
c.
16 m
d.
17 m


Solution:

Let n be the side of the square floor.

The area of a square = side × side.

256 = n × n
[Substitute the values.]

n2 = 256
[Multiply.]

n2 = 256
[Square root each side.]

n = 16
[Find the positive square root, since the side-length cannot be negative.]

The side-length of the square floor is 16 meters.


Correct answer : (3)
 8.  
What is the value of x, if 4x2 - 80 = 116?
a.
± 8
b.
± 10
c.
± 7
d.
± 9


Solution:

4x2 - 80 = 116
[Original equation.]

4x2 - 80 + 80 = 116 + 80
[Add 80 to both sides.]

4x2 = 196
[Simplify.]

x2 = 49
[Divide by 4 on both sides.]

x = ± 7
[Find the square roots.]


Correct answer : (3)
 9.  
A cow is tied with a rope in a meadow so that it can graze a maximum area of 78.50 ft2. What is the length of the rope?

a.
4 ft
b.
6 ft
c.
15 ft
d.
5 ft


Solution:

Let r be the length of the rope.

The maximum area that the cow can graze is the area of the circle, whose radius is equal to the length of the rope.

So, the maximum area that the cow can graze is π × (length of a rope)2
[Area of the circle = π × radius2.]

π × r2 = 78.50

r2 = 78.503.14 = 25
[Divide both sides by 3.14.]

r = 25
[Length of the rope is positive.]

= 5 × 5

= 5
[Simplify.]

So, the length of the rope is 5 ft.


Correct answer : (4)
 10.  
The area of a circular garden is given by the equation, A = 3.14r2, where A is the area of the garden in square ft and r is the radius of the garden in ft. What is the radius of the garden, if its area is 2826 square ft?
a.
35 ft
b.
300 ft
c.
30 ft
d.
15 ft


Solution:

Area of a circular garden, A = 3.14r2

2826 = 3.14r2
[Replace A with 2826.]

28263.14 = r2
[Divide each side by 3.14.]

900 = r2
[Simplify.]

r = 900 = ± 30
[Take square root on both sides.]

Radius of the garden cannot be negative. So, r = 30 ft.


Correct answer : (3)

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