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Pre-Algebra - Simple Adding and Subtracting Radicals

You might need more than an abacus for this quiz!

Pre-Algebra - Simple Adding and Subtracting Radicals

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.

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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. 

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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

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