Question

How do you simplify $\sqrt{5x-4}-\sqrt{x+8}=2$?

Answer 1

Step-1: Simplify the expression:

Rewrite the expression:

$\Rightarrow \sqrt{5x-4}=2+\sqrt{x+8}$

To remove the radical, square both sides of the equation:

$\Rightarrow {\left(\sqrt{5x-4}\right)}^2={\left(2+\sqrt{x+8}\right)}^2$

Cancel the square term with root:

$\Rightarrow 5x-4={\left(2+\sqrt{x+8}\right)}^2$

Step-2: Apply the formula:

Apply ${(\mathrm{a}+\mathrm{b})}^2$ formula on the right side:

$\begin{array}{rcl}5x-4& =& {\left(2+\sqrt{x+8}\right)}^2\\ & =& {\left(2\right)}^2+{\left(\sqrt{x+8}\right)}^2+2\left(2\right)\left(\sqrt{x+8}\right)\\ & =& 4+x+8+4\sqrt{x+8}\\ & =& 12+4\sqrt{x+8}+x\end{array}$

Rewrite the term:

$\begin{array}{rcl}& \Rightarrow & 12+4\sqrt{x+8}+x=5x-4\\ & \Rightarrow & 4\sqrt{x+8}=5x-x-4-12\\ & \Rightarrow & 4\sqrt{x+8}=4x-16\end{array}$

To remove the radical on the left side of the equation, square both sides of the equation:

$\Rightarrow {\left(4\sqrt{x+8}\right)}^2\begin{array}{rcl}& =& \end{array}{\left(4x-16\right)}^2$

Step-3: Use the power rule:

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to the term ${\left(4\sqrt{x+8}\right)}^2$:

$\begin{array}{rcl}& \Rightarrow & {\left({\begin{array}{l}4\left(x+8\right)\end{array}}^{\frac{1}{2}}\right)}^2={\left(4x-16\right)}^2\\ 4^2\times {\left({\left(x+8\right)}^{\frac{1}{2}}\right)}^2& =& {\left(4x-16\right)}^2\\ 16\left(x+8\right)& =& {\left(4x-16\right)}^2[\mathrm{Cancel}\mathrm{the}\mathrm{power}\mathrm{term}]\end{array}$

Step-4: Expand the expression:

$\begin{array}{rcl}& \Rightarrow & 16\left(x+8\right)={\left(4x-16\right)}^2\\ 16x+128& =& {\left(4x\right)}^2+{\left(16\right)}^2-2\left(4x\right)\left(16\right)\\ 16x+128& =& 16x^2+256-128x\end{array}$

Rewrite the terms:

$\begin{array}{rcl}& \Rightarrow & 16x^2+256-128x=16x+128\\ & \Rightarrow & 16x^2=16x+128-256+128x\\ & \Rightarrow & 16x^2=144x-128\end{array}$

Rewrite the equation:

$\begin{array}{rcl}& \Rightarrow & 16x^2-144x+128=0\end{array}$

Step-5: Factor the term:

$\begin{array}{rcl}& \Rightarrow & 16x^2-144x+128=0\end{array}$

Factor ${\displaystyle 16}$ out of ${\displaystyle 16x^2-144x+128}$:

$\Rightarrow 16(x^2-9x+8)=0$

Use the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$.

Here, whose product is ${\displaystyle 8}$ and whose sum is ${\displaystyle -9}$.

$\Rightarrow -8,-1$

Write the equation in factored form:

$\Rightarrow 16(x-8)(x-1)=0$

If any individual factor on the left side of the equation is equal to ${\displaystyle 0}$, the entire expression will be equal to ${\displaystyle 0}$:

$$\begin{gathered} \Rightarrow x-8=0 \\ \Rightarrow x-1=0 \end{gathered}$$

Hence, the values of $x$ are $\left(8,1\right)$.

Step-6: Exclude the solutions that do not make $\sqrt{\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{4}}\mathbf{-}\sqrt{\mathbf{x}\mathbf{+}\mathbf{8}}\mathbf{=}\mathbf{2}$ true:

Check for $x=1$:

$\begin{array}{rcl}& \Rightarrow & \sqrt{5\left(1\right)-4}-\sqrt{1+8}=2\\ \sqrt{1}-\sqrt{9}& =& 2\\ 1-3& =& 2\end{array}$

The value of $x=1$is not true.

Check for $x=8$:

$\begin{array}{rcl}& \Rightarrow & \sqrt{5\left(8\right)-4}-\sqrt{\left(8\right)+8}=2\\ \sqrt{36}-\sqrt{16}& =& 2\\ 6-4& =& 2\\ 2& =& 2\end{array}$

The value of $x=8$ is true.

Hence, the solution of the term $\sqrt{\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{4}}\mathbf{-}\sqrt{\mathbf{x}\mathbf{+}\mathbf{8}}\mathbf{=}\mathbf{2}$ is $\mathit{x}\mathbf{=}\mathbf{8}$.