Question

Solve each inequality algebraically.

$(x-3)(x+2)<x^2+3x+5$

Short answer

Solution set and the interval is:

$$\begin{gathered} \left\{x\right|x>-\frac{11}{4}\left\} \\ \right(-\frac{11}{4},\infty ) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero. 

Change the inequality equal to zero to make it easier and then get the value of x

First arrange the inequality,

$\begin{array}{r}(x-3)(x+2)<x^2+3x+5\\ x^2-x-6<x^2+3x+5\\ x^2-x-6-x^2-3x-5<0\\ -4x-11<0\end{array}$

Make it equal to zero,

$$\begin{gathered} -4x-11=0 \\ 4x+11=0 \\ x=-\frac{11}{4} \end{gathered}$$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(-\infty ,-\frac{11}{4})\cup (-\frac{11}{4},\infty )$

Step 3. Form the table

Since, $f\left(x\right)<0$, so is negative.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(-\infty ,-\frac{11}{4})$	-3	$1<0$
$(-\frac{11}{4},\infty )$	5	$-31<0$

Step 4. Solution set and interval

Since, only $-31<0$satisfy $f$. Thus,

role="math" localid="1646211136842" $$\begin{gathered} \left\{x\right|x>-\frac{11}{4}\left\} \\ \right(-\frac{11}{4},\infty ) \end{gathered}$$