Toolkit 69: Approaches to Solving Square Root Problems

Let

t=\sqrt{x-4}

Then

x=t^2+4

Now

x+4\sqrt{x-4} = t^2+4t+4 = (t+2)^2

and

x-4\sqrt{x-4} = t^2-4t+4 = (t-2)^2

Therefore, the equation becomes

|t+2|+|t-2|=4

Since

t\ge0

t+2+|t-2|=4

t+|t-2|=2

Case 1: $t\ge2$

t+t-2=2

2t=4

t=2=\sqrt{x-4}

x-4=4

x=8

Checking:

\begin{aligned} \sqrt{x+4\sqrt{x-4}} + \sqrt{x-4\sqrt{x-4}} &= \sqrt{8+4\sqrt4} + \sqrt{8-4\sqrt4}\\ &= \sqrt{16}+\sqrt0\\ &= 4 \end{aligned}

\boxed{x=8}\qquad\qquad\text{(1)}

Case 2: $0\le t<2$

t+|t-2| = t+2-t = 2

0\le t=\sqrt{x-4}<2

x-4<4

x<8

x\in[4,8)\qquad\qquad\text{(2)}

By (1) and (2),

\boxed{x\in[4,8]}