Chapter 1.9, Problem 8Q

Solution

Simplify the inequality $x^2-7x\ge 12$ .

The solution for the inequality is $x\in \left(-\infty ,\frac{1}{2}\left(7-\sqrt{97}\right)\right]\cup \left[\frac{1}{2}\left(7+\sqrt{97}\right),\infty \right)$ .

Given:

The inequality is $x^2-7x\ge 12$ .

Concept used:

If $u^n\ge A\forall \left(n\right)$ belongs to even integer then the solution of the inequality is written as:

  $x\le -{\left(A\right)}^{\frac{1}{n}}\text{ or }x\ge {\left(A\right)}^{\frac{1}{n}}\mathrm{......}\left(1\right)$

Calculation:

Simplify the inequality $x^2-7x\ge 12$ .

  $\begin{array}{c}{x}^2-7x\ge 12\mathrm{......}\left(\text{Given}\right)\\ x^2-7x+{\left(\frac{7}{2}\right)}^2\ge 12+{\left(\frac{7}{2}\right)}^2\mathrm{......}\left(\text{Add both sides by }{\left(\frac{7}{2}\right)}^2\right)\\ {\left(x-\frac{7}{2}\right)}^2\ge \frac{97}{4}\mathrm{......}\left(\text{Complete square}\right)\\ x-\frac{7}{2}\le -\sqrt{\frac{97}{4}}\text{ or }x-\frac{7}{2}\ge \sqrt{\frac{97}{4}}\mathrm{......}\left(\text{Using }\left(1\right)\right)\\ x\le \frac{7}{2}-\frac{\sqrt{97}}{2}\text{ or }x\ge \frac{7}{2}+\frac{\sqrt{97}}{2}\mathrm{......}\left(\text{Add }\frac{7}{2}\text{ to each side of inequality}\right)\\ x\in \left(-\infty ,\frac{1}{2}\left(7-\sqrt{97}\right)\right]\cup \left[\frac{1}{2}\left(7+\sqrt{97}\right),\infty \right)\end{array}$

Hence the solution for the inequality is $x\in \left(-\infty ,\frac{1}{2}\left(7-\sqrt{97}\right)\right]\cup \left[\frac{1}{2}\left(7+\sqrt{97}\right),\infty \right)$ .