Chapter 3.6, Problem 34E

Solution

To calculate: The value of x for the equation $x-6=\sqrt{3x}$ and also check the extraneous solution.

The resultant answer is $x=12$ and the extraneous solution is 3.

Given information:

The equation is $x-6=\sqrt{3x}$ .

Formula used:

Raise each side of the equation to the same power.

Calculation:

Write the original equation:

  $x-6=\sqrt{3x}$

Square each side of the equation,

  ${\left(x-6\right)}^2={\left(\sqrt{3x}\right)}^2$

Expand left side and simplify right side,

  $x^2+36-12x=3x$

Write the equation in standard form and factor the equation,

  $\begin{array}{c}{x}^2+36-12x-3x=0\\ x^2-15x+36=0\\ \left(x-12\right)\left(x-3\right)=0\end{array}$

Now apply zero-product property,

  $\begin{array}{c}x-12=0\text{or}x-3=0\\ x=12\text{or}x=3\end{array}$

Now check $x=12\text{or}x=3$ in the original equation.

  $x-6=\sqrt{3x}$

Now substitute 12 for x and simplify,

  $\begin{array}{c}12-6\stackrel{?}{=}\sqrt{3\cdot 12}\\ 6\stackrel{?}{=}\sqrt{36}\\ 6=6\end{array}$

Now substitute 3 for x and simplify,

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

Therefore, the only solution is 12 and the apparent solution 3 is extraneous.