Chapter 3.6, Problem 40E

Solution

To calculate: The value of x for the equation $x-2=\sqrt{\frac{3}{2}x-2}$ and also check the extraneous solution.

The resultant answer is $x=4$ and the extraneous solution is $\frac{3}{2}$ .

Given information:

The equation is $x-2=\sqrt{\frac{3}{2}x-2}$ .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  $x-2=\sqrt{\frac{3}{2}x-2}$

Square each side of the equation,

  $\begin{array}{c}{\left(x-2\right)}^2={\left(\sqrt{\frac{3}{2}x-2}\right)}^2\\ x^2-4x+4=\frac{3}{2}x-2\\ x^2-4x+6=\frac{3}{2}x\end{array}$

Multiply both sides by 2,

  $\begin{array}{c}2\left(x^2-4x+6\right)=\frac{3}{2}x\cdot 2\\ 2x^2-8x+12=3x\\ 2x^2-11x+12=0\end{array}$

Divide the equation into factors,

  $\begin{array}{c}2x^2-8x-3x+12=0\\ 2x\left(x-4\right)-3\left(x-4\right)=0\\ \left(x-4\right)\left(2x-3\right)=0\\ x=4\text{or}\frac{3}{2}\end{array}$

Now check $x=4\text{or}\frac{3}{2}$ in the original equation.

  $x-2=\sqrt{\frac{3}{2}x-2}$

Now substitute 4 for x and simplify,

  $\begin{array}{c}4-2\stackrel{?}{=}\sqrt{\frac{3}{2}\cdot 4-2}\\ 2\stackrel{?}{=}\sqrt{6-2}\\ 2=2\end{array}$

Now substitute $\frac{3}{2}$ for x and simplify,

  $\begin{array}{l}\frac{3}{2}-2\stackrel{?}{=}\sqrt{\frac{3}{2}\cdot \frac{3}{2}-2}\\ -\frac{1}{2}\stackrel{?}{=}\sqrt{\frac{9}{4}-2}\\ -\frac{1}{2}\stackrel{?}{=}\sqrt{\frac{1}{4}}\\ -\frac{1}{2}\ne \frac{1}{2}\end{array}$

Therefore, the only solution is 4 and the apparent solution $\frac{3}{2}$ is extraneous.