Answer
$x=4$
Work Step by Step
We are given:
$\sqrt{x}+1=\sqrt{11-\sqrt{x}}$
We square both sides:
$(\sqrt{x}+1)^{2}=(\sqrt{11-\sqrt{x}})^{2}$
$(\sqrt{x}+1)(\sqrt{x}+1)=(\sqrt{11-\sqrt{x}})(\sqrt{11-\sqrt{x}})$
$x+2\sqrt{x}+1=11-\sqrt{x}$
$x+2\sqrt{x}+1+\sqrt{x}=11$
$x+3\sqrt{x}+1=11$
$3\sqrt{x}=10-x$
We square both sides again:
$(3\sqrt{x})^{2}=(10-x)^{2}$
$9x=100-20x+x^{2}$
$100-29x+x^{2}=0$
$x^{2}-29x+100=0$
And factor:
$(x-4)(x-25)=0$
Equate each factor to zero, then solve each equation for $x$:
$(x-4)=0$ or $(x-25)=0$
$x=4$ or $x=25$
However, the solution $x=25$ does not work in the original equation:
$\sqrt{25}+1=\sqrt{11-\sqrt{25}}$
$5+1=\sqrt{11-5}$
$6=\sqrt{6}$
$36=6$
Since we got a false statement, the solution $x=25$ does not work.
Hence the only solution is $x=4$.
