Chapter 1.9, Problem 34E

Solution

To calculate: The value of x for the inequality $2x^2-6x-9\ge 11$ using a table.

The solution of the inequality is $\left(-\infty ,-2\right]\cup \left[5,\infty \right)$ .

Given information:

The inequality is $2x^2-6x-9\ge 11$ .

Formula used:

Substitute the values for x into the inequality.

Calculation:

Write the given inequality:

  $2x^2-6x-9\ge 11$

Subtract 11 from both sides to make the right side of the inequality equal to 0,

  $\begin{array}{c}2x^2-6x-9-11\ge 11-11\\ 2x^2-6x-20\ge 0\\ x^2-3x-10\ge 0\end{array}$

Substitute the $x=-4$ in the inequality,

  $\begin{array}{c}{\left(-4\right)}^2-3\left(-4\right)-10\ge 0\\ 16+12-10\ge 0\\ 18\ge 0\end{array}$

Substitute the $x=-3$ in the inequality,

  $\begin{array}{c}{\left(-3\right)}^2-3\left(-3\right)-10\ge 0\\ 9+9-10\ge 0\\ 8\ge 0\end{array}$

Substitute the $x=-2$ in the inequality,

  $\begin{array}{c}{\left(-2\right)}^2-3\left(-2\right)-10\ge 0\\ 4+6-10\ge 0\\ 0\ge 0\end{array}$

Similarly make a table by substituting the values of x in the inequality:

x	$-4$	$-3$	$-2$	$-1$	0	1	2	3	4	5
$x^2-2x-8$	18	8	0	$-6$	$-10$	$-12$	$-12$	$-10$	$-6$	0

The inequality $2x^2-6x-9\ge 11$ fails between $-2$ and 6 and true for all real numbers.

Therefore, $\left(-\infty ,-2\right]\cup \left[5,\infty \right)$ is the solutions for the inequality $2x^2-6x-9\ge 11$ .