Chapter 4.2, Problem 1QR

To determine

To solve: the given inequality.

Answer to Problem 1QR

  $\left(-\sqrt{3},\sqrt{3}\right)$ .

Explanation of Solution

Given:

  $2x^2-6<0$ .

Calculation:

Consider,

  $2x^2-6<0$

By adding 6 on both sides, it follows that

  $\begin{array}{c}2x^2-6+6<0+6\\ 2x^2<6\end{array}$

Dividing by 2 on both sides, it follows that

  $\begin{array}{c}\frac{2x^2}{2}<\frac{6}{2}\\ x^2<3\\ -\sqrt{3}<x<\sqrt{3}\end{array}$

So, the solution is $\left(-\sqrt{3},\sqrt{3}\right)$ .