Answer
$x= 0$ and $2.31$
Work Step by Step
Given equation-
$1+\sqrt x$ = $\sqrt {1+x^{2}}$ , in the interval [-1,5]
Substituting $\sqrt x$ = z and hence $x^{2}$ = $z^{4}$
$1+z$ = $\sqrt {1+z^{4}}$,
Now squaring on both the sides-
$1+z^{2} +2z $ = $1+z^{4}$
i.e. $z^{4}-z^{2}-2z$ = $0$
Graphing $y$ = $z^{4}-z^{2}-2z$, using graphing calculator-
We get, $z= 0$ and $1.52$
Now $x$ = $z^{2}$
Therefore, $x$ = $0$ and $1.52^{2}$
i.e. $x$ = $0$ and $2.31$
We are asked to find all solutions 'x' that satisfy $-1 \leq x \leq 5$, Therefore in the interval [-1,5]-
$x$ = $0$ and $2.31$