How do you solve #sqrt(50)+sqrt(2)# ?

1 Answer
Sep 9, 2015

Answer:

You can simplify #sqrt(50)+sqrt(2) = 6sqrt(2)#

Explanation:

If #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)# and #sqrt(a^2) = a#

So:

#sqrt(50)+sqrt(2) = sqrt(5^2*2)+sqrt(2) = sqrt(5^2)sqrt(2) + sqrt(2)#

#= 5sqrt(2)+1sqrt(2) = (5+1)sqrt(2) = 6sqrt(2)#

In general you can try to simplify #sqrt(n)# by factorising #n# to identify square factors. Then you can move the square roots of those square factors out from under the square root.

e.g. #sqrt(300) = sqrt(10^2*3) = 10sqrt(3)#