Exercise 44 Page 286

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 will start by solving the related equation. We see above that $a=\text{-}3,$ $b=\text{-}10,$ and $c=1.$ Let's substitute these values into the Quadratic Formula to solve the equation. Now we can calculate the first root using the positive sign and the second root using the negative sign.

$x=\frac{10\pm \sqrt{112}}{\text{-}6}$
$x=\frac{10+\sqrt{112}}{\text{-}6}$	$x=\frac{10-\sqrt{112}}{\text{-}6}$
$x=\text{-}\frac{5}{3}-\frac{2\sqrt{7}}{3}$	$x=\text{-}\frac{5}{3}+\frac{2\sqrt{7}}{3}$
$x\approx \text{-}3.43$	$x\approx 0.097$

Step $2$

The solutions of the related equation are approximately $\text{-}3.43$ and $0.097.$ Let's plot them on a number line. Since the original is a strict inequality, the points will be open.

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, $x<\text{-}3.43.$ For simplicity, we will choose $x=\text{-}4.$ Since $x=\text{-}4$ did not produce a true statement, the interval $x<\text{-}3.43$ is not part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
$\text{-}3.43<x<0.097$	$0$	$0>\text{-}1$ $✓$	Yes
$x>0.097$	$1$	$\text{-}13\ngtr \text{-}1$ $\times$	No

We can now write the solution set. Finally, we can show the solution set on a number line.