Question

Question #55801

Answer 1

`(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`

Answer 2

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`