Chapter 4, Problem 3CR

To determine

The solution of the inequality $x^2-3x<4$ and graph the solution set of the inequality

Answer to Problem 3CR

Solution:

The solution of the inequality is $\left(-1,4\right)$ . The graph of the solution set is

  

Explanation of Solution

Given information:

The inequality $x^2-3x<4$

Rearrange the inequality, so that $0$ is on the right side

  $x^2-3x-4<0$ .

The above inequality is equivalent to the original inequality

Now, graph the function $f\left(x\right)=x^2-3x-4$ to find where $f\left(x\right)<0$

Set $f\left(x\right)=x^2-3x-4=0$

Here, the two numbers whose sum is $b=-3$ and whose product is $a\cdot c=-4$ are $-4,1$

So, rewrite the equation as

  $x^2+x-4x-4=0$

  $\Rightarrow x\left(x+1\right)-4\left(x+1\right)=0$

  $\Rightarrow \left(x+1\right)\left(x-4\right)=0$

  $\Rightarrow x+1=0$ or $x-4=0$

  $\Rightarrow x=-1,x=4$ .

The $x$ -intercept of the graph is $x=-1,x=4$ .

Now, $f\left(0\right)=0^2-3\times 0-4=-4$

Hence, $y=-4$ is the $y$ -intercept of the graph

The function $f\left(x\right)$ is the parabola with vertex is at $x=-\frac{b}{2a}=-\frac{-3}{2\times 1}=3/2$

  $f\left(\frac{3}{2}\right)={\left(\frac{3}{2}\right)}^2-3\times \frac{3}{2}-4=\frac{9}{4}-\frac{9}{2}-4=-\frac{25}{4}$ .

Thus, the vertex of the parabola is $\left(\frac{3}{2},-\frac{25}{4}\right)$

Hence, the graph of the function $f\left(x\right)=x^2-3x-4$

  

Thus, the graph below the $x\text{-axis}$ is in the set $\left(-1,4\right)$

Hence, the graph of the solution set is