Chapter 1.9, Problem 62E

Solution

Simplify the inequality $-3x^2+4x-5\le 2$ by any method.

The solution for the inequality doesn’t exist.

Given:

The inequality is $-3x^2+4x-5\le 2$ .

Concept used:

If $u^n\ge A\forall A$ belongs to positive integer, then $u\in \left(-\infty ,-A^{\frac{1}{n}}\right]\cup \left[A^{\frac{1}{n}},\infty \right)$ and if $\left(A\right)$ belongs to negative integer then no real solution exists for the inequality.

Calculation:

Simplify the inequality $-3x^2+4x-5\le 2$ algebraically.

  $\begin{array}{c}-3x^2+4x-5\le 2\mathrm{......}\left(\text{Given}\right)\\ 3x^2-4x+5\ge -2\mathrm{......}\left(\text{Multiply by }-1\right)\\ 3x^2-4x\ge -7\\ x^2-\frac{4}{3}x\ge -\frac{7}{3}\mathrm{......}\left(\text{Divide by 3 both sides}\right)\\ x^2-\frac{4}{3}x+{\left(\frac{2}{3}\right)}^2\ge -\frac{7}{3}+{\left(\frac{2}{3}\right)}^2\mathrm{......}\left(\text{Add }{\left(\frac{2}{3}\right)}^2\text{both sides}\right)\\ {\left(x-\frac{2}{3}\right)}^2\ge -\frac{17}{9}\\ {\left(x-\frac{2}{3}\right)}^2\ge -\frac{17}{9}\mathrm{......}\left(\begin{array}{l}\text{From the concept stated above,}\\ \text{ no real solution will exist}\end{array}\right)\\ x\notin R\mathrm{......}\left(\text{where }R\text{is any real number}\right)\end{array}$

Hence the solution for the inequality doesn’t exist.