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_ ___
2Ö5 + 5Ö125
You can't break 5 down because it's prime, but you can
break down 125 into primes
125 = 5·25 = 5·5·5, so substitute 5·5·5 for 125
_ _____
2Ö5 + 5Ö5·5·5
Since it's a square root we group like factors into
twos. (If it were a cube root you would group into
threes)
_ ________
2Ö5 + 5Ö(5·5)·5
Since (5·5) is the same as 5², so we can substitute
5² for (5·5)
____
2V5 + 5Ö5²·5
Now we can take individual square roots of the factors
under the second radical
_ __ _
2Ö5 + 5Ö5²·Ö5
Taking the square root of '5 squared' just takes
away the square and the radical, and you just have 5:
_ _
2Ö5 + 5·5·Ö5
_ _
2Ö5 + 25Ö5
_
Now you can factor out the Ö5
_
Ö5(2 + 25)
2 plus 25 is 27
_
Ö5(27)
Write the 27 first
_
27Ö5
Edwin