How do you simplify #(sqrt(15)+sqrt(35)+sqrt(21)+5)/(sqrt(3)+2sqrt(5)+sqrt(7))# to get #(sqrt(7)+sqrt(3))/2# ?

1 Answer
George C.
Sep 1, 2017

See explanation...

Explanation:

Given:

#(sqrt(15)+sqrt(35)+sqrt(21)+5)/(sqrt(3)+2sqrt(5)+sqrt(7))#

We could rationalise the denominator by multiplying both numerator and denominator by:

#(sqrt(3)+2sqrt(5)-sqrt(7))(sqrt(3)-2sqrt(5)+sqrt(7))(sqrt(3)-2sqrt(5)-sqrt(7))#

...but since we have been given what is at least supposed to be the answer, we can work backwards instead...

Note that:

#(sqrt(7)+sqrt(3))(sqrt(3)+2sqrt(5)+sqrt(7))#

#=sqrt(7)(sqrt(3)+2sqrt(5)+sqrt(7))+sqrt(3)(sqrt(3)+2sqrt(5)+sqrt(7))#

#=sqrt(21)+2sqrt(35)+7+3+2sqrt(15)+sqrt(21)#

#=2(sqrt(15)+sqrt(35)+sqrt(21)+5)#

So, dividing both ends by #2(sqrt(3)+2sqrt(5)+sqrt(7))# we get:

#(sqrt(15)+sqrt(35)+sqrt(21)+5)/(sqrt(3)+2sqrt(5)+sqrt(7)) = (sqrt(7)+sqrt(3))/2#

Related questions
See all questions in Square Roots and Irrational Numbers
Impact of this question
7340 views around the world
You can reuse this answer
Creative Commons License