US UKIndia

Every Question Helps You Learn

You might need more than an abacus for this quiz!

This Math quiz is called 'Pre-Algebra - Simple Adding and Subtracting Radicals' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.

It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us

In order to add radicals, there must be “like” terms in the problem. A problem with like terms would be 5x + 10x. Both terms contain the variable “x”. So the solution to 5x + 10x = 15x.

For each radical expression shown in the questions below, add or subtract to find the solution.

Go straight to Quiz

1.
36√9 - 17√9 =
19√18
19√9
19√81
36-17√9
36√9 - 17√9 has like terms, i.e., √9 so the coefficient numbers 36 and 17 can be subtracted or 36 - 17 = 19 and the like term is added.
Solution: 19√9
Answer (b) is the correct solution
2.
10√48 + 15√3
-5√3
55√3
25√51
40√3
10√48 + 15√3
As we do not have like terms we must simplify as follows:
10√48 = 10√16 ● 3
As 16 is a perfect square in that 4 ● 4 = 16, the 4 goes before or to the left of the √ symbol giving you 10 ● 4√3
10 ● 4 = 40 and then add on the √3 giving you 40√3. Now the problem can be rewritten as: 40√3 + 15√3 = 55√3
Solution: 55√3
Answer (b) is the correct solution
3.
199√xy - 99√xy
100√xy
xy
√100xy
100√xy2
199√xy - 99√xy has like terms, i.e., √xy so the coefficient numbers 199 and 99 can be subtracted or 199 - 99 = 100 and the like term is added.
Solution: 100√xy
Answer (a) is the correct solution
4.
6√125 - 4√5
2√5
26√25
26√5
-24√5
6√125 - 4√5
As we do not have like terms we must simplify as follows:
6√125 = 6√25 ● 5
As 25 is a perfect square in that 5 ● 5 = 25, the 5 goes before or to the left of the √ symbol giving you 6 ● 5√5
6 ● 5 = 30 and then add on the √5 giving you 30√5. Now the problem can be rewritten as: 30√5 - 4√5 = 26√5
Solution: 26√5
Answer (c) is the correct solution
5.
45√72 - 221√2
-178√70
-178√72 - 2
49√2
49√62
45√72 - 221√2
As we do not have like terms we must simplify as follows:
45√72 = 45√36 ● 2
As 36 is a perfect square in that 6 ● 6 = 36, the 6 goes before or to the left of the √ symbol giving you 6 ● 45√2
6 ● 45 = 270 and then add on the √2 giving you 270√2. Now the problem can be rewritten as: 270√2 - 221√2 = 49√2
Solution: 49√2
Answer (c) is the correct solution
6.
√144 - √64
√-80
-√80
4√
8√12
√144 - √64
As we do not have like terms we must simplify as follows:
√144 has a square root as 12 ● 12 = 144. Therefore, the square 12 is then placed before or to the left of the √ symbol giving us 12√
Next, √64 also has a square root as 8 ● 8 = 64. The square 8 is then placed before or to the left of the √ symbol giving us 8√
We now have like terms of √ and the problem can be rewritten as:
12√ - 8√ = 4√
Solution: 4√
Answer (c) is the correct solution
7.
12√12 + 23√12 =
35√12 + 12
35√12
12 + 23√12 + 12
12 + 23√12
12√12 + 23√12 has like terms, i.e., √12 so the coefficient numbers 12 and 23 can be added or 12 + 23 = 35 and the like term is added.
Solution: 35√12
Answer (b) is the correct solution
8.
5√3 + 9√3 =
14√ 3 + 3
5 + 9√32
14√6
14√3
5√3 + 9√3 has like terms, i.e., √3 so the coefficient numbers 5 and 9 can be added or 5 + 9 = 14 and the like term is added.
Solution: 14√3
Answer (d) is the correct solution
9.
28√x + 31√x
59√x
59√x2
59√x + x
59
28√x + 31√x has like terms, i.e., √x so the coefficient numbers 28 and 31 can be added or 28 + 31 = 59 and the like term is added.
Solution: 59√x
Answer (a) is the correct solution
10.
5apq + 7apq
12a2pq
12apq2
12a2pq2
12apq
5apq + 7apq has like terms, i.e., a and√pq so the coefficient numbers 5 and 7 can be added or 5a + 7a = 12a and the like term is added.
Solution: 12apq
Answer (d) is the correct solution
Author:  Christine G. Broome