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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.
[readmore]Let’s look at the next square root problem of 6√2 + 8√2. The like term here is √2. In addition to the √2 being like terms, so are the coefficient numbers “6” and “8” so you get to add those together, i.e., 6 + 8 = 14. The 14 is then placed before √2 and the solution is 14√2.
The same rule holds true for subtraction.
There must be “like” terms in the problem. Knowing this, if we look at the problem of: 9√p  3√p the like term is √p and the coefficient numbers are 9  3 = 6. The 6 is then placed before the like term of √p giving us the solution of 6√p.
Now let’s look at another problem. 10√12 + 2√3. You cannot add these binomials because the terms are not like terms. However, you can simplify 10√12 to read 10√4 ● 3 and then the number 4 is the perfect square root of 2 so the 2 moves to the left of the √ symbol so that the problem now looks like: 10 ● 2√3. Next, 10 ● 2 = 20 giving you 20√3. Now we go back to the original problem and rewrite is as 20√3 + 2√3. We have √3 as our like term and then 20 + 2 = 22 which is placed before our like term giving us the solution of 22√3.
[/readmore]For each radical expression shown in the questions below, add or subtract to find the solution.















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 

√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 
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 

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 
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 

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 
5a√pq + 7a√pq 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: 12a√pq Answer (d) is the correct solution 

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 
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 
Solution: 14√3
Answer (d) is the correct solution