McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 6 Page 219

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

{ x|0.75 ≤ x ≤ 4}

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. 4x^2-19x=- 12 ⇕ 4x^2+( - 19)x+ 12=0We see above that a= 4, b= - 19, and c= 12. Let's substitute these values into the Quadratic Formula to solve the equation.
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 19)±sqrt(( - 19)^2-4( 4)( 12))/2( 4)
â–Ľ
Simplify right-hand side
x=19±sqrt((- 19)^2-4(4)(12))/2(4)
x=19±sqrt(361-4(4)(12))/2(4)
x=19±sqrt(361-192)/8
x=19±sqrt(169)/8
x = 19 ± 13/8
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x = 19 ± 13/8
x = 19 + 13/8 x = 19 - 13/8
x = 32/8 x = 6/8
x = 4 x = 0.75

Step 2

The solutions of the related equation are 4 and 0.75. 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 ≤ 0.75. For simplicity, we will choose x=- 1.
4x^2-19x ≤ - 12
4( - 1)^2-19( - 1)? ≤- 12
â–Ľ
Simplify left-hand side
4(1)-19(- 1)? ≤- 12
4+19? ≤- 12
23 ≰ - 12
Since x=- 1 did not produce a true statement, the interval x ≤ 0.75 is not part of the solution. Similarly, we can test the other two intervals.
Interval Test Value Statement Is It Part of the Solution?
0.75 ≤ x ≤ 4 1 - 15 ≤ - 12 ✓ Yes
x ≥ 4 5 5 ≰ - 12 * No

We can now write the solution set and show it on a number line. { x|0.75 ≤ x ≤ 4} or [ 0.75, 4 ]