Question

Solve each inequality algebraically.

$x^2+3x\ge 10$

Short answer

Solution set and the interval is:

$$\begin{gathered} \left\{x\right|x\le -5orx\ge 2\left\} \\ \right(-\infty ,-5]\cup [2,\infty ) \end{gathered}$$

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.

$$\begin{gathered} x^2+3x=10 \\ {x}^2+3x-10=0 \\ (x+5)(x-2)=0 \\ x=-5;x=2 \end{gathered}$$

Step 2. Form the intervals

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

So the interval we have is:

$(-\infty ,-5)\cup (-5,2)\cup (2,\infty )$

Step 3. Form the table

Since, $f\left(x\right)\ge 0$, so is positive or zero.

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

Interval	Number chosen	Resultant
$(-\infty ,-5)$	-6	8$\ge$0
$(-5,2)$	1	-6$\ge$0
$(2,\infty )$	4	18$\ge$0

Step 4. Solution set and interval

Since,

$8\ge 0,18\ge 0$satisfy $f$. Thus,

role="math" localid="1646198622225" $$\begin{gathered} \left\{x\right|x\le -5orx\ge 2\left\} \\ \right(-\infty ,-5]\cup [2,\infty ) \end{gathered}$$