Q. 13 - Precalculus Enhanced with Graphing Utilities

Question

In problems 7-22, solve the inequality.     

$x^2+x>12$

Step-by-Step Solution

Answer

The solution set is $(-\infty ,-4) or (3,\infty )$.

Step 1. Given Information

The given inequality is $x^2+x>12$.

Step 2. Find the intercepts and vertex

• Rewrite the given inequality.

$x^2+x-12>0$

• The value of $f\left(0\right)=-12$.
• So, the y-intercept is $(0,-12)$.
• Factor the equation.

$\begin{array}{rcl}{x}^2+x-12& =& 0\\ x^2+4x-3x-12& =& 0\\ x(x+4)-3(x+4)& =& 0\\ (x-3)(x+4)& =& 0\end{array}$

• Equate the factors to 0.

$\begin{array}{rcl}x-3& =& 0\\ x& =& 3\\ x+4& =& 0\\ x& =& -4\end{array}$

• So, the x-intercepts are $(3,0),(-4,0)$.
• The vertex of the parabola exists at $\frac{-b}{2a}=\frac{-1}{2\left(1\right)}=\frac{-1}{2}$.
• Find the value of the function at $x=-\frac{1}{2}$.

$\begin{array}{rcl}f(-\frac{1}{2})& =& {(-\frac{1}{2})}^2+(-\frac{1}{2})-12\\ & =& \frac{1}{4}-\frac{1}{2}-12\\ & =& \frac{1-2-48}{4}\\ & =& \frac{-49}{4}\end{array}$

• So, the vertex is $(-\frac{1}{2},\begin{array}{rcl}& & \frac{-49}{4}\end{array})$.

Step 3. Plot the graph

Plot the parabola on a graph using the vertex and intercepts calculated in the previous step.  

Step 4. Find the solution set

• From the graph, it is observed that for x belonging to $(-\infty ,-4) or (3,\infty )$, $y>0$.
• So, the solution set of the given inequality is $(-\infty ,-4) or (3,\infty )$.