Answer
$x-intercept = 2$
The graph of equation $y= \sqrt (x+2)+\sqrt (x-1)-3$ is shown in Graph$(c)$
Work Step by Step
$y= \sqrt (x+2)+\sqrt (x-1)-3$
The $y$ coordinate corresponding to an $x$-intercept is zero. So, to find $x$ -intercept,
$ \sqrt (x+2)+\sqrt (x-1)-3 = 0$
$ \sqrt (x+2)+\sqrt (x-1) = 3$
Squaring on both sides,
$ (\sqrt (x+2)+\sqrt (x-1))^{2}= 3^{2}$
Using $(A+B)^{2}=A^{2}+2AB+B^{2}$
$ (\sqrt (x+2))^{2}+ 2\sqrt (x+2)\sqrt (x-1)+ (\sqrt (x-1))^{2} = 9$
$ (x+2)+ 2\sqrt (x+2)\sqrt (x-1)+ (x-1) = 9$
$ 2x+1+ 2\sqrt (x+2)\sqrt (x-1)= 9$
$ 2x+ 2\sqrt (x+2)\sqrt (x-1)= 8$
$ 2(x+ \sqrt (x+2)\sqrt (x-1))= 8$
$ x+ \sqrt (x+2)\sqrt (x-1)= 4$
$ \sqrt (x+2)\sqrt (x-1)= 4-x$
Again, squaring on both sides,
$ (\sqrt (x+2)\sqrt (x-1))^{2}= (4-x)^{2}$
Using $(A-B)^{2}=A^{2}-2AB+B^{2}$
$ (x+2)(x-1)= 16-8x+x^{2}$
$ x^{2}+2x-x-2= 16-8x+x^{2}$
Combine like terms.
$x-2=16-8x$
$9x=18$
$x=2$
The graph of equation $y= \sqrt (x+2)+\sqrt (x-1)-3$ is shown in Graph$(c)$
