Chapter 7.1, Problem 31WE

To determine

To Solve:

  $0.6x^2+2=2.4x$ by completing the square.

Answer to Problem 31WE

  $x=2\pm \frac{\sqrt{6}}{3}$

Explanation of Solution

Given:

The equation is :

  $0.6x^2+2=2.4x$

Calculation:

Given:

  $\begin{array}{l}0.6x^2+2=2.4x\\ \Rightarrow \frac{6}{10}{x}^2+2=\frac{24}{10}x\\ \Rightarrow x^2+\frac{10}{3}=4x\mathrm{..............}[multiply\text{ each side by }\frac{\text{10}}{6}\text{ to make the coefficient of square term as 1]}\\ \Rightarrow x^2-4x=-\frac{10}{3}\mathrm{.................}[\text{subtract (4x+}\frac{10}{3}\text{) from each side to isolate the constant]}\\ \Rightarrow x^2-4x+{\left(\frac{-4}{2}\right)}^2=-\frac{10}{3}+{\left(\frac{-4}{2}\right)}^2\mathrm{....}\left[\begin{array}{l}\text{add the square of the half of the coefficient}\\ \text{of x each side}\end{array}\right]\\ \Rightarrow x^2-4x+4=-\frac{10}{3}+4\\ \Rightarrow {\left(x-2\right)}^2=\frac{2}{3}\mathrm{.................................}[{\left(a-b\right)}^2=a^2-2ab+b^2]\\ \Rightarrow x-2=\pm \sqrt{\frac{2}{3}}\mathrm{.................................}[\text{take square root each side and }\sqrt{-1}=i\text{]}\\ \Rightarrow x=2\pm \frac{\sqrt{2}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\mathrm{....................................}[add\text{ 2 each side and rationalize the denominator]}\\ \Rightarrow x=2\pm \frac{\sqrt{6}}{3}\end{array}$

So,

  $x=2\pm \frac{\sqrt{6}}{3}$