yes this is my homework... v.v..

1) √63 + √175 + √112
2) √28x + √63x
3) √45 + √180
4) √52 - √1300
5) 5√98 - 3√32
6) √32 + √128
7) √147 + 6√3
8) √168 + √42
9) 5√17 + 17√5
10) 6√3 + √300
11) -2√3b + √27b
12) 4√2m + 6√3m - 4√2m
13) √50m + √72m
14) √16z + 2√8z - 3√z
15) √216t + √96t
16) 4√52x + √117x - 2√13
17) 3√96k + 2√180

You need to factor the radicand of each radical down to the prime factorization. If you get a pair of factors, pull it outside the radicand as a coefficient. Add the coefficients.

I'll try it, I get lost when it comes to coefficients

Try the first problem, and post the answer here. I'll let you know if you've done it correctly or not.

√63 + √175 + √112

√3*21 √5*35 √4*28

3√21 5√35 4√28


Am I right so far?. I don't know

You're right so far, but you need the prime factorization. You should end up with the same radicand, so they can be added.

Let me complete the first one for you, so you can see how it's done:

We have √63 + √175 + √112

√63 = √3 X 3 X 7. There's a pair of 3's and a 7 left over, so pull out the 3 to make 3√7

√175 = √5 X 5 X 7. There's a pair of 5's and a 7 left over, so pull out the 5. 5√7

√112 = √2 X 2 X 2 X 2 X 7. We have 2 pairs of 2's, so that becomes 4. 4√7

3 + 5 + 4 = 12. So the final answer is 12√7

So I just need to find the prime factorization of the radicands?

Yes, that's all you need to do. Pair up any pairs of factors and pull them out, and anything left over (singles)would be left under the radicand.

Oh OK, well thank you