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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 10 Page 210

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

{x|x≠ 7}

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^2-14x=- 49 ⇔ 1x^2+( - 14)x+ 49=0We see above that a= 1, b= - 14, and c= 49. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 14)±sqrt(( - 14)^2-4( 1)( 49))/2( 1)

▼

Simplify right-hand side

- (- a)=a

x=14±sqrt((- 14)^2-4(1)(49))/2(1)

Calculate power

x=14±sqrt(196-4(1)(49))/2(1)

Multiply

x=14±sqrt(196-196)/2

Subtract term

x=14±sqrt(0)/2

Calculate root

x=14 ± 0/2

Identity Property of Addition

x=14/2

Calculate quotient

x=7

The root of the equation is 7

Step 2

The solution of the related equation is 7. Let's plot it on a number line. Since the original is a strict inequality, the point 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 < 7. For simplicity, we will choose x=0.

x^2-14x >- 49

x= 0

( 0)^2-14( 0)? >- 49

▼

Simplify left-hand side

Calculate power

0-14(0)? >- 49

Zero Property of Multiplication

0-0? >- 49

Identity Property of Addition

0 > - 49

Since x=0 produced a true statement, the interval x < 7 is a part of the solution. Similarly, we can test the other interval.

Interval	Test Value	Statement	Is It Part of the Solution?
x > 7	10	- 40 > - 49 ✓	Yes

We can now write the solution set and show it on a number line. {x|x≠ 7} or (- ∞ , 7) ⋃ (7, + ∞)

Subchapter links

Exercises p.210-213