Chapter 12.2, Problem 25WE

To determine

To Solve: The equation $x+1=\frac{2+x}{4x}$ and write irrational roots in simplest radical form.

Answer to Problem 25WE

The required solution set is $\left\{\frac{-3+\sqrt{41}}{8},\frac{-3-\sqrt{41}}{8}\right\}$ .

Explanation of Solution

Given information:

  $x+1=\frac{2+x}{4x}$

Calculation:

  $\begin{array}{l}x+1=\frac{2+x}{4x}\\ 4x^2+4x=2+x\\ 4x^2+3x=2\end{array}$

(Rewrite the equation)

  $x^2+\frac{3x}{4}=\frac{1}{2}$

(Divide both sides by 4 so that the coefficient of $x^2$ will be 1)

  $x^2+\frac{3x}{4}+{\left(\frac{3}{8}\right)}^2=\frac{1}{2}+{\left(\frac{3}{8}\right)}^2$

(Add square of half the coefficient of x to both sides)

  $\begin{array}{l}{x}^2+\frac{3x}{4}+\left(\frac{9}{64}\right)=\frac{1}{2}+\left(\frac{9}{64}\right)\\ {\left(x+\frac{3}{8}\right)}^2=\frac{41}{64}\end{array}$

(Simplify)

  $\begin{array}{l}x+\frac{3}{8}=\sqrt{\frac{41}{64}}\\ x+\frac{3}{8}=\pm \frac{\sqrt{41}}{8}\end{array}$

(Perform square root on both sides)

  $x=-\frac{3}{8}\pm \frac{\sqrt{41}}{8}$

Therefore,

  $\begin{array}{l}x=-\frac{3}{8}+\frac{\sqrt{41}}{8}\\ x=\frac{-3+\sqrt{41}}{8}\end{array}$

Or

  $\begin{array}{l}x=-\frac{3}{8}-\frac{\sqrt{41}}{8}\\ x=\frac{-3-\sqrt{41}}{8}\end{array}$

Therefore, the solution set is $\left\{\frac{-3+\sqrt{41}}{8},\frac{-3-\sqrt{41}}{8}\right\}$ .