Q. 338 - Intermediate Algebra

Question

In the following exercises, solve.
$\sqrt{3x+1}=1+\sqrt{2x-1}$

Step-by-Step Solution

Answer

The solutions of the given equation are $x=1$ and $x=5$.

Step 1. Given information.

The given equation is:

$\sqrt{3x+1}=1+\sqrt{2x-1}$

Step 2. Simplify the equation.

Take square on both sides.

$\begin{array}{rcl}(\sqrt{3x+1}{)}^2& =& (1+\sqrt{2x-1}{)}^2\\ 3x+1& =& 1+2\sqrt{2x-1}+(2x-1)\\ 3x+1-2x& =& 2\sqrt{2x-1}\\ x+1& =& 2\sqrt{2x-1}\end{array}$

Step 3. Solve the equation.

Square both sides.

$\begin{array}{rcl}(x+1{)}^2& =& 4(2x-1)\\ x^2+2x+1-8x+4& =& 0\\ x^2-6x+5& =& 0\end{array}$

Splitting the middle term, we get

$\begin{array}{rcl}{x}^2-5x-x+5& =& 0\\ x(x-5)-1(x-5)& =& 0\\ (x-1)(x-5)& =& 0\\ x& =& 1,5\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $x=1$ and $x=5$.