Question #55801

2 Answers
Apr 25, 2016

Answer:

#(sqrt(x+sqrt(x-4))+sqrt(x-4sqrt(x-4)))^2=4x-16#

Explanation:

#x=a#

#x^2=a^2#

#4sqrt(x-4)=b#

#16x-64=b^2#

#(sqrt(x+sqrt(x-4))+sqrt(x-4sqrt(x-4)))^2=(sqrt(a+b)+sqrt(a-b))^2#

#(sqrt(a+b)+sqrt(a-b))^2=a+cancel(b)+2*sqrt(a+b)*sqrt(a-b)+a-cancel(b)#

#(sqrt(a+b)+sqrt(a-b))^2=2a+2sqrt((a+b)(a-b))#

#(a+b)(a-b)=a^2-b^2#

#(sqrt(a+b)+sqrt(a-b))^2=2a+2sqrt(a^2-b^2)#

#(sqrt(a+b)+sqrt(a-b))^2=2x+2sqrt(x^2-16x+64)#

#(sqrt(a+b)+sqrt(a-b))^2=2x+2sqrt((x-8)^2)#

#(sqrt(a+b)+sqrt(a-b))^2=2x+2(x-8)#

#(sqrt(a+b)+sqrt(a-b))^2=2x+2x-16#

#(sqrt(a+b)+sqrt(a-b))^2=4x-16#

#(sqrt(x+sqrt(x-4))+sqrt(x-4sqrt(x-4)))^2=4x-16#

Apr 25, 2016

Answer:

Slightly different approach in the beginning to that of Ali Ergin but we end up at the same point part way through.

#4x-64#

Explanation:

Let #a=4sqrt(x-4)# giving

#(sqrt(x+a)+sqrt(x-a)color(white)(.))^2#

#(sqrt(x+a)color(white)(.))^2 + 2sqrt(x+a)sqrt(x-a)+(sqrt(x-a)color(white)(.))^2#

#x+cancel(a)+ 2sqrt(x+a)sqrt(x-a)+x-cancel(a)#

#2x+2sqrt(x+a)sqrt(x-a)#

#2x+2sqrt((x+a)(x-a))#

#2x+2sqrt(x^2-a^2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

But #a=4sqrt(x-4) => a^2=16(x-4)=16x-64#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So by substitution for a giving

#2x+2sqrt(x^2-(16x-64))#

#2x+2sqrt(x^2-16x+64))#

Note that #(-8)xx(-8)=+64" and "-8-8=-16#

#2x+2sqrt((x-8)^2)#

#2x+2(x-8)#

#4x-64#