What is the the value of #sqrt(6+sqrt(20))#?

The answer has to be expressed in the form of surds.

1 Answer
NickTheTurtle
Dec 29, 2017

#sqrt(6+sqrt(20))=1+sqrt(5)#

Explanation:

Here is one way to solve it.

Assume that #sqrt(6+sqrt(20))=a+sqrt(b)# where #a# and #b# are nonnegative integers.

Then, squaring both sides, #6+sqrt(20)=a^2+2asqrt(b)+b#. Equating coefficients by the rationality of the terms, we find
#{(a^2+b=6),(2asqrt(b)=sqrt(20)=2sqrt(5)):}#

From the second equation, we have #a^2b=5#. Multiply both sides of the first equation by #b# to get #a^2b+b^2=6b#, or #b^2-6b+5=(b-5)(b-1)=0#.

The solutions of this quadratic equation are #b=1# or #5#, but, when #b=1#, #a=sqrt(5)#.

Thus, the only solution for integers #a# and #b# is #a=1,b=5#.

So, we have #sqrt(6+sqrt(20))=1+sqrt(5)#.

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