McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 5 Page 219

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

{ x| x < - 5 or x>- 1 }

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. x^2+6x = - 5 ⇔ 1x^2+ 6x+ 5=0We see above that a= 1, b= 6, and c= 5. 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)( 5))/2( 1)
Simplify right-hand side
x=- 6±sqrt(36-4(1)(5))/2(1)
x=- 6±sqrt(36-20)/2
x=- 6±sqrt(16)/2
x = - 6 ± 4/2
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x = - 6 ± 4/2
x = - 6 + 4/2 x = - 6 - 4/2
x = - 2/2 x = - 10/2
x = - 1 x = - 5

Step 2

The solutions of the related equation are - 1 and - 5. 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 < - 5. For simplicity, we will choose x=- 6.
x^2+6x > - 5
( - 6)^2+6( - 6)? >- 5
Simplify left-hand side
36+6(- 6) > - 5
36-36> - 5
0 > - 5
Since x=- 6 produced a true statement, the interval x < - 5 is part of the solution. Similarly, we can test the other two intervals.
Interval Test Value Statement Is It Part of the Solution?
- 5 < x < - 1 - 2 - 8 ≯ - 5 * No
x > - 1 0 0 > - 5 ✓ Yes

We can now write the solution set and show it on a number line. { x| x < - 5 or x>- 1 } or (- ∞, - 5) ⋃ (- 1, ∞)