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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 39 Page 211

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

{ x| x ≤ - 1.58 or x ≥ 1.58 }

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+4=9 ⇔ 2x^2+ 0x+( - 5)=0We see above that a= 2, b= 0, and c= - 5. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 0±sqrt(0^2-4( 2)( - 5))/2( 2)

▼

Simplify right-hand side

Calculate power

x=0±sqrt(0-4(2)(- 5))/2(2)

Multiply

x=0±sqrt(0+40)/4

Identity Property of Addition

x=±sqrt(40)/4

SplitIntoFactors

Split into factors

x=±sqrt(4* 10)/4

sqrt(a* b)=sqrt(a)*sqrt(b)

x=±sqrt(4)sqrt(10)/4

Calculate root

x=± 2sqrt(10)/4

a/b=.a /2./.b /2.

x=± sqrt(10)/2

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

x=±sqrt(10)/2
x=sqrt(10)/2	x=- sqrt(10)/2
x≈ 1.58	x≈ - 1.58

Step 2

The solutions of the related equation are approximately - 1.58 and 1.58. 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 ≤ - 1.58. For simplicity, we will choose x=- 2.

2x^2+4 ≥ 9

x= - 2

2( - 2)^2+4? ≥ 9

▼

Simplify left-hand side

Calculate power

2(4)+4? ≥ 9

Multiply

8+4? ≥ 9

Add terms

12 ≥ 9 ✓

Since x=- 2 produced a true statement, the interval x ≤ - 1.58 is a part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
- 1.58 ≤ x ≤ 1.58	0	4 ≱ 9 *	No
x ≥ 1.58	2	12 ≥ 9 ✓	Yes

We can now write the solution set and show it on a number line. { x| x ≤ - 1.58 or x ≥ 1.58 } or (- ∞, - 1.58] ⋃ [1.58, ∞)

Subchapter links

Exercises p.210-213