Exercise 12 Page 285

To solve the quadratic inequality algebraically, we will follow three steps.

1. Solve the related quadratic equation.
2. Plot the solutions on a number line.
3. Test a value from each interval to see if it satisfies the original inequality.

Step $1$

We see above that $a=1,$ $b=\text{-}4,$ and $c=\text{-}21.$ Let's substitute these values into the Quadratic Formula to solve the equation.

$x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}$

SubstituteValues

Substitute values

$x=\frac{\text{-}(\text{-}4)\pm \sqrt{(\text{-}4{)}^2-4(1)(\text{-}21)}}{2(1)}$

▼

Simplify right-hand side

NegNeg

$\text{-}(\text{-}a)=a$

$x=\frac{4\pm \sqrt{(\text{-}4{)}^2-4(1)(\text{-}21)}}{2(1)}$

CalcPow

Calculate power

$x=\frac{4\pm \sqrt{16-4(1)(\text{-}21)}}{2(1)}$

Multiply

Multiply

$x=\frac{4\pm \sqrt{16+84}}{2}$

AddTerms

Add terms

$x=\frac{4\pm \sqrt{100}}{2}$

CalcRoot

Calculate root

$x=\frac{4\pm 10}{2}$

Now we can calculate the first root using the positive sign and the second root using the negative sign.

$x=\frac{4\pm 10}{2}$
$x=\frac{4+10}{2}$	$x=\frac{4-10}{2}$
$x=\frac{4}{2}+\frac{10}{2}$	$x=\frac{4}{2}-\frac{10}{2}$
$x=7$	$x=\text{-}3$

Step $2$

The solutions of the related equation are $\text{-}3$ and $7.$ Let's plot them on a number line. Since the original is not a strict inequality, the points will be closed.

Step $3$

Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval, $x\le \text{-}3.$ For simplicity, we will choose $x=\text{-}4.$

$x^2-4x\le 21$

Substitute

$x=\text{-}4$

$(\text{-}4{)}^2-4(\text{-}4)\stackrel{?}{\le }21$

▼

Simplify left-hand side

CalcPow

Calculate power

$16-4(\text{-}4)\stackrel{?}{\le }21$

Multiply

Multiply

$16+16\stackrel{?}{\le }21$

AddTerms

Add terms

$32\nleqq 21\times$

Since $x=\text{-}4$ did not produce a true statement, the interval $x\le \text{-}3$ is not part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
$\text{-}3\le x\le 7$	$0$	$0\le 21$ $✓$	Yes
$x\ge 7$	$10$	$60\nleqq 21$ $\times$	No

We can now write the solution set and show it on a number line.

$$\begin{array}{c}\{x|\text{-}3\le x\le 7\}\text{or}[\text{-}3,7]\end{array}$$