McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 44 Page 219

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

All Real Numbers

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. 6x + x^2 ≥ - 9 ⇔ 1x^2+ 6x+ 9 ≥ 0We see above that a= 1, b= 6, and c= 9. 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)( 9))/2( 1)
â–Ľ
Simplify right-hand side
x=- 6±sqrt(36-4(1)(9))/2(1)
x=- 6±sqrt(36-36)/2
x=- 6±sqrt(0)/2
x= -6/2
x= -3

Step 2

The solution of the related equation is - 3. Let's plot it on a number line. Since the original is not a strict inequality, the point 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 ≤ - 3. For simplicity, we will choose x=- 4.
6x+x^2 ≥ - 9
6( - 4)+ ( - 4)^2 ? ≥ - 9
â–Ľ
Simplify left-hand side
6(- 4)+ 16 ? ≥ - 9
-24 + 16 ? ≥ - 9
-8 ≥ - 9 ✓
Since x=- 4 produced a true statement, the interval x ≤ - 3 is part of the solution. Similarly, we can test the other interval.
Interval Test Value Statement Is It Part of the Solution?
x ≤ - 3 -4 - 8 ≥ -9 ✓ Yes
x ≥ -3 0 0 ≥ - 9 ✓ Yes

We can now write the solution set and show it on a number line. x ≤ -3 and x≥ - 3 Since this inequality is not strict, -3 is included in the solution set.

We see above that the solution set is all real numbers.