## Precalculus (6th Edition)

$\color{blue}{\left\{\dfrac{-29}{2} - \dfrac{\sqrt{1241}}{2}, -\dfrac{29}{2}+\dfrac{\sqrt{1241}}{2}\right\}}$
Square both sides of the equation: $$(\sqrt{x}+1)^2=(\sqrt{11-\sqrt{x}})^2 \\(\sqrt{x})^2+2(\sqrt{x})(1)+1^2=11-\sqrt{x} \\x+2\sqrt{x}+1=11-\sqrt{x}$$ Add $\sqrt{x}$ and subtract $1$ to both sides of the equation: $$x+2\sqrt{x}+1+\sqrt{x}-1=11-\sqrt{x}+\sqrt{x}-1 \\x+3\sqrt{x}=10$$ Subtract $x$ to both sides of the equation to obtain: $$3\sqrt{x}=10-x$$ Square both sides of the equation to obtain: $$(3\sqrt{x})^2=(10-x)^2 \\9(x)=10^2-2(10)(x)+x^2 \\9x=100-20x+x^2$$ Subtract $9x$ to both sides of the equation: $$9x-9x=100-20x-9x \\0=100-29x-x^2 \\x^2+29x-100=0$$ The quadratic equation above has $a=1, b=29$, and $c=-100$. Solve the equation using the quadratic formula to obtain: $$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\x=\dfrac{-29\pm \sqrt{29^2-4(1)(-100)}}{2(1)} \\x=\dfrac{-29 \pm \sqrt{841+400}}{2} \\x=\dfrac{-29\pm \sqrt{1241}}{2}$$ Thus, the solution set is: $\color{blue}{\left\{\dfrac{-29}{2} - \dfrac{\sqrt{1241}}{2}, -\dfrac{29}{2}+\dfrac{\sqrt{1241}}{2}\right\}}$.