Q. 8.127 - Intermediate Algebra

Question

Solve: $3-\sqrt{x}=\sqrt{x-3}$

Step-by-Step Solution

Answer

The solution is $x=4$.

Step 1. Square both sides and simplify.

Consider the equation $3-\sqrt{x}=\sqrt{x-3}$.

The radical on the right is isolated. Now we square both sides and simplify.

$\begin{array}{rcl}{(3-\sqrt{x})}^2& =& {\left(\sqrt{x-3}\right)}^2 \\ 3^2-2·3·\sqrt{x}+{\left(\sqrt{x}\right)}^2& =& x-3\\ 9-6\sqrt{x}+x& =& x-3\end{array}$

Step 2. Isolate the radical and solve for x

Now, subtract $x+9$ from both sides

$\begin{array}{rcl}9-6\sqrt{x}+x-(x+9)& =& x-3-(x+9)\\ 9-6\sqrt{x}+x-x-9& =& x-3-x-9\\ -6\sqrt{x}& =& -12\end{array}$

Now divide both sides by $-6$

$\begin{array}{rcl}\frac{-6\sqrt{x}}{-6}& =& \frac{-12}{-6}\\ \sqrt{x}& =& 2\end{array}$

Now, square both sides.

$\begin{array}{rcl}{\left(\sqrt{x}\right)}^2& =& 2^2\\ x& =& 4\end{array}$

Step 3. Check the solution

Substitute $4$ for $x$ in the original equation.

$\begin{array}{rcl}3-\sqrt{x}& =& \sqrt{x-3}\\ 3-\sqrt{4}& =& \sqrt{4-3}\\ 3-2& =& \sqrt{1}\\ 1& =& 1\end{array}$

As the statement is true, so the found solution is correct.