Chapter 8.2, Problem 6QR

Solution

To find: The solution of the given equation algebraically.

The solution of $\sqrt{6x+12}-\sqrt{4x+9}=1$ is $x=4$ .

Given information: The equation is $\sqrt{6x+12}-\sqrt{4x+9}=1$ .

Calculation:

Simplify the equation $\sqrt{6x+12}-\sqrt{4x+9}=1$ as follows.

Rewrite the given equation as follows.

  $\begin{array}{c}\sqrt{6x+12}-\sqrt{4x+9}=1\\ \sqrt{4x+9}=1-\sqrt{6x+12}\end{array}$

Take square root on both sides.

  $\begin{array}{c}{\left(\sqrt{4x+9}\right)}^2={\left(1-\sqrt{6x+12}\right)}^2\\ 4x+9=\left(1^2-2\cdot 1\cdot \sqrt{6x+12}+{\left(\sqrt{6x+12}\right)}^2\right)\\ 4x+9=1-2\sqrt{6x+12}+6x+12\\ 2\sqrt{6x+12}=2x+4\end{array}$

Again, take square root on both sides.

  $\begin{array}{c}{\left(2\sqrt{6x+12}\right)}^2={\left(2x+4\right)}^2\\ 4\left(6x+12\right)=4x^2+16x+16\\ 24x+48=4x^2+16x+16\\ 4x^2-8x-32=0\end{array}$

Factor the $4x^2-8x-32=0$ as follows.

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

Check the root $x=-2$ is true or false as follows.

  $\begin{array}{r}\sqrt{6\left(-2\right)+12}-\sqrt{4\left(-2\right)+9}\stackrel{?}{=}1\\ 0-1\stackrel{?}{=}1\\ -1\ne 1\end{array}$

That is, the root $x=-2$ is not a solution of $\sqrt{6x+12}-\sqrt{4x+9}=1$ .

Check the root $x=4$ is true or false as follows.

  $\begin{array}{r}\sqrt{6\left(4\right)+12}-\sqrt{4\left(4\right)+9}\stackrel{?}{=}1\\ 6+5\stackrel{?}{=}1\\ 1=1\end{array}$

That is, the root $x=4$ is solution of $\sqrt{6x+12}-\sqrt{4x+9}=1$ .

Therefore, the solution of $\sqrt{6x+12}-\sqrt{4x+9}=1$ is $x=4$ .