### Home > INT3 > Chapter 4 > Lesson 4.2.2 > Problem4-56

4-56.

Denae and Gina are trying to solve $\sqrt{x+5}+\sqrt{5}=5$. They know they have to square both sides of the equation to get rid of the radicals, but when they try it, they get $x+5+2\sqrt{x(x+5)}+x=25$, which does not look much easier to deal with. Gina has an idea.

"Let’s start by putting the radicals on opposite sides of the equation and then squaring.” she says. Her work is shown at right. $\left. \begin{array}{l}{ \sqrt { x + 5 } + \sqrt { x } = 5 }\\ \ \ \ \ \ \ \ \ \ \ \ { \sqrt { x + 5 } = 5 - \sqrt { x } }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { x + 5 = 25 - 10 \sqrt { x } + x }\end{array} \right.$
“But there’s still a radical!” she exclaims, disappointed.

“That’s okay,” says Denae, “there’s only one radical now, so we can get the term with the radical by itself and then square everything again.” Homework Help ✎

$x+5=25-10\sqrt{x}+x$

$-20=-10\sqrt{x}$

$2=\sqrt{x}$

$?=x$

2. Use Denae and Gina’s method to solve $\sqrt{2x-2}-\sqrt{x}=1$.

$\sqrt{2x-2}=1+\sqrt{x}$

$\left(\sqrt{2x-2}\right)^{2}=\left(1+\sqrt{x}\right)^2$

$2x-2=1+1\sqrt{x}+1\sqrt{x}+x$

$x-3=2\sqrt{x}$

$\left(x-3\right)^2=\left(2\sqrt{x}\right)^2$

$x^2-3x-3x+9=4x$

Continue solving. Note that you will need to factor and use the Zero Product Property.
Be sure to check both of your answers in the original equation.