Chapter 7.1, Problem 39WE

To determine

Solve the equation $\sqrt{x-4}-\frac{2}{\sqrt{x-4}}=1$ and check for extraneous roots by substitution of answer into the original equation.

Answer to Problem 39WE

The solution set for the equation is $\left\{8\right\}$ .

Explanation of Solution

Given information: The given equation is,

  $\sqrt{x-4}-\frac{2}{\sqrt{x-4}}=\mathrm{1....}\left(1\right)$

Calculation: Simplify (1) ,

  $\begin{array}{c}\sqrt{x-4}-\frac{2}{\sqrt{x-4}}=\mathrm{1....}\left(\text{Given}\right)\\ x-4+\frac{4}{x-4}-4=\mathrm{1....}\left(\text{Squared}\right)\\ x+\frac{4}{x-4}=9\\ x^2-4x+4=9x-\mathrm{36....}\left(\begin{array}{l}\text{Multiply}\left(x-4\right)\\ \text{ both sides}\end{array}\right)\end{array}$

Further solving,

  $\begin{array}{c}{x}^2-13x+40=0\\ \left(x-5\right)\left(x-8\right)=0\\ x-5=0\text{or}x-8=0\\ x=5\text{ or}x=8\end{array}$

Since the root $x=5$ does not satisfy (1) as it makes the left hand side of given equation (1) equal to $-1$ whereas right hand side is equal to $1$ so neglected. And substitute for $x=8$ in the equation (1) to check the result.

  $\begin{array}{c}\sqrt{\left(8\right)-4}-\frac{2}{\sqrt{\left(8\right)-4}}=\mathrm{1....}\left(1\right)\mathrm{....}\left(\text{From }\left(1\right)\right)\\ \sqrt{4}-\frac{2}{\sqrt{4}}=1\\ 2-\frac{2}{2}=1\\ 1=\mathrm{1....}\left(\text{which is true}\right)\end{array}$

Since $x=8$ satisfy (1) therefore the only solution for the given equation is $x=8$ . So, the solution set for the given equation is $\left\{8\right\}$ .