Chapter 12.5, Problem 20WE

To determine

Solve the quadratic equation.

Answer to Problem 20WE

  $x=\frac{9\pm \sqrt{129}}{12}$

Explanation of Solution

Given:

The equation is :

  $\frac{1}{3x}=\frac{2x-3}{2}$

Calculation:

The given equation $\frac{1}{3x}=\frac{2x-3}{2}$ .

So,

  $\begin{array}{l}\frac{1}{3x}=\frac{2x-3}{2}\\ \Rightarrow 2=6x^2-9x\mathrm{.............}[\text{multiply each side by 6x]}\\ \Rightarrow 6x^2-9x-2=\mathrm{0.......}[\text{subtract 2 from each side]}\\ \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-(-9)\pm \sqrt{{(-9)}^2-4(6)(-2)}}{2(6)}\\ \Rightarrow x=\frac{9\pm \sqrt{129}}{12}\end{array}$