Chapter 5.7, Problem 3Q

Solution

To solve: The equation and check for extraneous solutions.

The solution for the equation is $x=0.$

Given Information:

The equation is $\frac{3x+1}{x^2-2}=\frac{x+1}{x-2}$ .

Calculation:

  $\frac{3x+1}{x^2-2}=\frac{x+1}{x-2}$

Apply cross multiplication:

  $\begin{array}{c}\left(3x+1\right)\left(x-2\right)=\left(x^2-2\right)\left(x+1\right)\\ 3x^2-5x-2=x^3+x^2-2x-2\\ x^3+x^2-2x-2+2=3x^2-5x-2+2\\ x^3+x^2-2x=3x^2-5x\\ x^3+x^2-2x+5x=3x^2-5x+5x\\ x^3+x^2+3x=3x^2\\ x^3+x^2+3x-3x^2=3x^2-3x^2\\ x^3-2x^2+3x=0\\ x\left(x^2-2x+3\right)=0\\ x=0\text{or}{x}^2-2x+3=0\end{array}$

Thus,

  $\begin{array}{c}{x}_{1,2}=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\cdot 1\cdot 3}}{2\cdot 1}\\ x_{1,2}=\frac{-\left(-2\right)\pm 2\sqrt{2}i}{2\cdot 1}\\ x=1+\sqrt{2}i,x=1-\sqrt{2}i\end{array}$

Now, check for extraneous solutions:

At $x=0:$

  $\begin{array}{l}\frac{3\left(0\right)+1}{0^2-2}=\frac{0+1}{0-2}\\ -\frac{1}{2}=-\frac{1}{2}\\ \text{True}\end{array}$

Hence, the solution of the equation is $x=0$ .