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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

5. Quadratic Inequalities

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Exercise 49 Page 212

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

{ x| x < - 1.06 or x > 7.06 }

Practice makes perfect

To solve the quadratic inequality algebraically, we will follow three steps.

Solve the related quadratic equation.
Plot the solutions on a number line.
Test a value from each interval to see if it satisfies the original inequality.

Step 1

We will start by solving the related equation. - 2x^2+12x=- 15 ⇔ - 2x^2+ 12x+ 15=0We see above that a= - 2, b= 12, and c= 15. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 12±sqrt(12^2-4( - 2)( 15))/2( - 2)

▼

Simplify right-hand side

Calculate power

x=- 12±sqrt(144-4(- 2)(15))/2(- 2)

Multiply

x=- 12±sqrt(144+120)/- 4

Add terms

x=- 12±sqrt(264)/- 4

Now we can calculate the first root using the positive sign and the second root using the negative sign.

x=- 12±sqrt(264)/- 4
x=- 12 + sqrt(264)/- 4	x=- 12 - sqrt(264)/- 4
x=- 12/- 4+sqrt(264)/- 4	x=- 12/- 4-sqrt(264)/- 4
x=3-sqrt(264)/4	x=3+sqrt(264)/4
x≈ - 1.06	x≈ 7.06

Step 2

The solutions of the related equation are approximately - 1.06 and 7.06. 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 < - 1.06. For simplicity, we will choose x=- 2.

- 2x^2+12x < - 15

x= - 2

- 2( - 2)^2+12( - 2)? <- 15

▼

Simplify left-hand side

Calculate power

- 2(4)+12(- 2)? <- 15

Multiply

- 8-24? <- 15

Subtract term

- 32 < - 15 ✓

Since x=- 2 produced a true statement, the interval x < - 1.06 is a part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
- 1.06 < x < 7.06	0	0 ≮ - 15 *	No
x > 7.06	10	- 80 < - 15 ✓	Yes

We can now write the solution set and show it on a number line. { x| x < - 1.06 or x > 7.06 } or (- ∞, - 1.06) ⋃ (7.06, ∞)

Subchapter links

Exercises p.210-213