Chapter 1.6, Problem 41E

To determine

To calculate: The solution of the equation, $2\sqrt{x+1}-\sqrt{2x+3}=1$ and check the solution

The solution of the given equation, $2\sqrt{x+1}-\sqrt{2x+3}=1$ is $x=3$.

Calculation:

Consider the provided equation,

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

Isolate the radical.

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

Square each side and solve.

$\begin{array}{c}{\left(2\sqrt{x+1}\right)}^2={\left(1+\sqrt{2x+3}\right)}^2\\ 4\left(x+1\right)=1+2x+3+2\sqrt{2x+3}\\ 4x+4=1+2x+3+2\sqrt{2x+3}\\ x=\sqrt{2x+3}\end{array}$

Again, square each side.

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

Write the above equation in standard form.

$x^2-2x-3=0$

Now factorize the above equation.

$\left(x-3\right)\left(x+1\right)=0$

Put the first factor equal to zero.

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

Put the second factor equal to zero.

$\begin{array}{c}x+1=0\\ x=-1\end{array}$

Check,

Put $x=3,-1$ in the equation, $2\sqrt{x+1}-\sqrt{2x+3}=1$.

First put $x=3$.

$\begin{array}{r}2\sqrt{3+1}-\sqrt{2\left(3\right)+3}\stackrel{?}{=}1\\ 2\sqrt{4}-\sqrt{9}\stackrel{?}{=}1\\ 4-3\stackrel{?}{=}1\\ 1=1\end{array}$

Which is true.

Now put $x=-1$.

$\begin{array}{r}2\sqrt{-1+1}-\sqrt{2\left(-1\right)+3}\stackrel{?}{=}1\\ 2\sqrt{0}-\sqrt{1}\stackrel{?}{=}1\\ \pm 1\ne 1\end{array}$

Which is false, and therefore $x=-1$ is an extraneous solution.

Hence, the solution of the given equation is $x=3$.