Chapter 4.6, Problem 56AYU

In Problems 49-60, solve each inequality algebraically.

$\left(x\text{ - 3}\right)\left(x\text{ + 2}\right)\text{ < }{x}^2\text{ + 3}x\text{ + 5}$

To determine

To solve: The inequality algebraically.

Answer to Problem 56AYU

$\left\{x|-\frac{11}{4}<x<\infty \right\}$

Explanation of Solution

Given:

$\left(x-3\right)\left(x+2\right)<x^2+3x+5$

Calculation:

$\left(x-3\right)\left(x+2\right)<x^2+3x+5$

$x^2-x-6<x^2+3x+5$

$-4x<11$

$4x+11>0$

$x>-\frac{11}{4}$

The solution set is $\left\{x|-\frac{11}{4}<x<\infty \right\}$ or, using interval notation $\left(-\frac{11}{4},\infty \right)$.