McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 9 Page 210

Solve the related quadratic equation, plot the solutions on a number line, and test a value from each interval.

{ x| - 8 < x < 2 }

Practice makes perfect

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. 1x^2+ 6x+( - 16)=0We see above that a= 1, b= 6, and c= - 16. Let's substitute these values into the Quadratic Formula to solve the equation.
x=- b±sqrt(b^2-4ac)/2a
x=- 6±sqrt(( 6)^2-4( 1)( - 16))/2( 1)
â–Ľ
Simplify right-hand side
x=- 6±sqrt(36-4(1)(- 16))/2(1)
x=- 6±sqrt(36+64)/2
x=- 6±sqrt(100)/2
x=- 6 ± 10/2
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x=- 6 ± 10/2
x=- 6 + 10/2 x=- 6 - 10/2
x=- 6/2 + 10/2 x=- 6/2 - 10/2
x=2 x=- 8

Step 2

The solutions of the related equation are - 8 and 2. 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 < - 8. For simplicity, we will choose x=- 10.
x^2+6x-16 < 0
( - 10)^2+6( - 10)-16? <0
â–Ľ
Simplify left-hand side
100+6(-10)-16? <0
100-60-16? <0
24 ≮ 0
Since x=- 10 did not produce a true statement, the interval x < - 8 is not part of the solution. Similarly, we can test the other two intervals.
Interval Test Value Statement Is It Part of the Solution?
- 8 < x < 2 0 - 16 < 0 âś“ Yes
x > 2 3 11 ≮ 0 * No

We can now write the solution set and show it on a number line. { x| - 8 < x < 2 } or (- 8, 2)