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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 37 Page 211

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

{ x| x ≤ - 2.32 or x ≥ 4.32 }

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. - x^2+2x=- 10 ⇔ - 1x^2+ 2x+ 10=0We see above that a= - 1, b= 2, and c= 10. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 2±sqrt(2^2-4( - 1)( 10))/2( - 1)

▼

Simplify right-hand side

Calculate power

x=- 2±sqrt(4-4(- 1)(10))/2(- 1)

Multiply

x=- 2±sqrt(4+40)/- 2

Add terms

x=- 2±sqrt(44)/- 2

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

x=- 2±sqrt(44)/- 2
x=- 2 + sqrt(44)/- 2	x=- 2 - sqrt(44)/- 2
x=- 2/- 2+sqrt(44)/- 2	x=- 2/- 2-sqrt(44)/- 2
x=1-sqrt(44)/2	x=1+sqrt(44)/2
x≈ - 2.32	x≈ 4.32

Step 2

The solutions of the related equation are approximately - 2.32 and 4.32. 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 ≤ - 2.32. For simplicity, we will choose x=- 4.

- x^2+2x ≤ - 10

x= - 4

- ( - 4)^2+2( - 4)? ≤- 10

▼

Simplify left-hand side

Calculate power

- 16+2(- 4)? ≤- 10

Multiply

- 16-8? ≤- 10

Subtract terms

- 24 ≤ - 10 ✓

Since x=- 4 did produced a true statement, the interval x ≤ - 2.32 is a part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
- 2.32 ≤ x ≤ 4.32	0	0 ≰ - 10 *	No
x ≥ 4.32	5	- 15 ≤ - 10 ✓	Yes

We can now write the solution set and show it on a number line. { x| x ≤ - 2.32 or x ≥ 4.32 } or (- ∞, - 2.32] ⋃ [4.32, ∞)

Subchapter links

Exercises p.210-213