Answer
$x$-intercept = 5
The graph of equation $y= \sqrt (x-4)+\sqrt (x+4)-4$ is shown in Graph$(a)$
Work Step by Step
$y= \sqrt (x-4)+\sqrt (x+4)-4$
To find $x$ -intercept, $y=0$
$ \sqrt (x-4)+\sqrt (x+4)-4 =0$
$ \sqrt (x-4)+\sqrt (x+4) =4$
Squaring on both sides,
$ (\sqrt (x-4)+\sqrt (x+4))^{2} =4^{2}$
Using $(A+B)^{2}=A^{2}+2AB+B^{2}$
$x-4+2\sqrt (x-4)\sqrt (x+4) + x+4 = 16$
Combine like terms.
$2x+2\sqrt (x-4)\sqrt (x+4) = 16$
$2(x+\sqrt (x-4)\sqrt (x+4)) = 16$
Divide both sides by $2$
$x+\sqrt (x-4)\sqrt (x+4) = 8$
$\sqrt (x-4)\sqrt (x+4) = 8-x$
Again, squaring on both sides,
$(\sqrt (x-4)\sqrt (x+4))^{2} = (8-x)^{2}$
$ (x-4)(x+4) = (8-x)^{2}$
Using $(a-b)(a+b)=a^{2}-b^{2}$ and $(A-B)^{2}=A^{2}-2AB+B^{2}$
$x^{2}-16=64-16x+x^{2}$
$x^{2}+16x-x^{2}=64+16$
$16x=80$
$x=5$
The graph of equation $y= \sqrt (x-4)+\sqrt (x+4)-4$ is shown in Graph$(a)$
