Chapter 8.2, Problem 9QR

Solution

To find: The exact solution of the given equation by completing the square.

The solution of $2x^2-6x-3=0$ is $\frac{-\sqrt{15}+3}{2}\text{ and }\frac{\sqrt{15}+3}{2}$ .

Given information: The equation is $2x^2-6x-3=0$ .

Calculation:

Rewrite the given equation as follows:

  $\begin{array}{c}2x^2-6x-3=0\\ 2\left(x^2-3x-\frac{3}{2}\right)=0\end{array}$

Compare the equation $x^2+\left(-3x\right)+\left(-\frac{3}{2}\right)=0$ with $x^2+ax+a^2=0$ then the value of $2a=-3$ and hence $a=-\frac{3}{2}$ .

Add and subtract the number ${\left(-\frac{3}{2}\right)}^2$ with $x^2+\left(-3x\right)+\left(-\frac{3}{2}\right)=0$ as follows:

  $\begin{array}{c}2\left(x^2+\left(-3x\right)+\left(-\frac{3}{2}\right)-{\left(-\frac{3}{2}\right)}^2+{\left(-\frac{3}{2}\right)}^2\right)=0\\ 2\left({\left(x+\left(-\frac{3}{2}\right)\right)}^2+\left(-\frac{3}{2}\right)-{\left(-\frac{3}{2}\right)}^2\right)=0\\ 2\left({\left(x-\frac{3}{2}\right)}^2-\frac{3}{2}-\frac{9}{4}\right)=0\end{array}$

On further simplifications.

  $\begin{array}{c}2\left({\left(x-\frac{3}{2}\right)}^2-\frac{6}{4}-\frac{9}{4}\right)=0\\ 2\left({\left(x-\frac{3}{2}\right)}^2-\frac{15}{4}\right)=0\\ 2{\left(x-\frac{3}{2}\right)}^2-\frac{15}{2}=0\end{array}$

Solve the equation $2{\left(x-\frac{3}{2}\right)}^2-\frac{15}{2}=0$ as follows:

  $\begin{array}{c}2{\left(x-\frac{3}{2}\right)}^2=\frac{15}{2}\\ {\left(x-\frac{3}{2}\right)}^2=\frac{15}{4}\\ x-\frac{3}{2}=\pm \frac{\sqrt{15}}{2}\end{array}$

Simplify further:

  $\begin{array}{c}x=\pm \frac{\sqrt{15}}{2}+\frac{3}{2}\\ x=\pm \frac{\sqrt{15}+3}{2}\\ x=\frac{-\sqrt{15}+3}{2}\text{ and }\frac{\sqrt{15}+3}{2}\end{array}$

Therefore, the solution of $2x^2-6x-3=0$ is $\frac{-\sqrt{15}+3}{2}\text{ and }\frac{\sqrt{15}+3}{2}$ .