How do you simplify #(2sqrt(7)+35)/[sqrt(7)]#?

1 Answer
Nov 4, 2015

Answer:

#2+5sqrt7#

Explanation:

It would help if there was something that we could cancel out. At first glance, there is a #sqrt7# in the numerator and the denominator, so lets see if we can do something about that. If we factor #35# we get;

#35=5*7#

But the #7# can be further factored by taking the square root.

#7=sqrt7^2=sqrt7*sqrt7#

So #35# becomes;

#35=5*sqrt7*sqrt7#

Now we can start simplifying the expression.

#(2sqrt7 + 35)/sqrt7=(2sqrt7 + 5*sqrt7*sqrt7)/sqrt7#

Move the #sqrt7# out of the numerator.

#(sqrt7(2+5sqrt7))/sqrt7#

Now the #sqrt7#s cancel and we have;

#(cancelsqrt7(2+5sqrt7))/cancelsqrt7=2+5sqrt7#