Chapter 3.6, Problem 49E

Solution

To calculate: The value of x for the equation $\sqrt{x+2}=2-\sqrt{x}$ and also check the extraneous solution.

The required solution is $x=\frac{1}{2}$ .

Given information:

The equation is $\sqrt{x+2}=2-\sqrt{x}$ .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  $\sqrt{x+2}=2-\sqrt{x}$

Square each side of the equation and simplify,

  $\begin{array}{c}{\left(\sqrt{x+2}\right)}^2={\left(2-\sqrt{x}\right)}^2\\ x+2=4+x-4\sqrt{x}\end{array}$

Isolate the radical expression,

  $4\sqrt{x}=2$

Divide each side by 4,

  $\begin{array}{c}\frac{4\sqrt{x}}{4}=\frac{2}{4}\\ \sqrt{x}=\frac{1}{2}\end{array}$

Square each side again and simplify,

  $\begin{array}{c}{\left(\sqrt{x}\right)}^2={\left(\frac{1}{2}\right)}^2\\ x=\frac{1}{4}\end{array}$

Now check $x=\frac{1}{4}$ in the original equation.

  $\sqrt{x+2}=2-\sqrt{x}$

Now substitute $\frac{1}{4}$ for x and simplify,

  $\begin{array}{c}\sqrt{\frac{1}{4}+2}\stackrel{?}{=}2-\sqrt{\frac{1}{4}}\\ \sqrt{\frac{9}{4}}\stackrel{?}{=}2-\frac{1}{2}\\ \frac{3}{2}=\frac{3}{2}\end{array}$

Therefore the solution is $\frac{1}{2}$ .