MHB How Do You Solve and Graph This Quadratic Inequality?

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To solve the quadratic inequality (x^2 - 1)/(x^2 + 8x + 15) ≥ 0, critical points are found by setting the numerator and denominator to zero, yielding x = -1, 1 for the numerator and x = -3, -5 for the denominator. The number line is divided into intervals based on these points, specifically (-∞, -5), (-5, -3), (-3, -1), (-1, 1), and (1, ∞). Testing these intervals shows that the solution includes (-∞, -5) and (1, ∞), while x = -3 and x = -5 are excluded due to being undefined. The final solution is confirmed as (-∞, -5) U (1, ∞).
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Section 2.6
Question 50Solve the quadratic inequality.

(x^2 - 1)/(x^2 + 8x + 15) ≥ 0

x^2 - 1 = (x - 1)(x + 1)

Question:

Do I solve the numerator or denominator to find the end points?

Setting each numerator factor to 0 we get x = 1, x = -1.

Factor denominator.

x^2 + 8x + 15 = (x + 3)(x + 5)

Set each denominator factor to 0.

x + 3 = 0

x = -3

x + 5 = 0

x = -5

Question:

Do I place x = -3 & x = -5 on the number line to test each interval?

Is the correct number line as follows?

<-----(-5)-----(-3)-----(-1)-----(1)----->

It looks like we must test 5 intervals, correct?
 
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RTCNTC said:
Section 2.6
Question 50Solve the quadratic inequality.

(x^2 - 1)/(x^2 + 8x + 15) ≥ 0

x^2 - 1 = (x - 1)(x + 1)

Question:

Do I solve the numerator or denominator to find the end points?

You want to determine the roots of both the numerator and denominator as your critical numbers (i.e., values for which the expression can change sign).

RTCNTC said:
Setting each numerator factor to 0 we get x = 1, x = -1.

Factor denominator.

x^2 + 8x + 15 = (x + 3)(x + 5)

Set each denominator factor to 0.

x + 3 = 0

x = -3

x + 5 = 0

x = -5

Question:

Do I place x = -3 & x = -5 on the number line to test each interval?

Is the correct number line as follows?

<-----(-5)-----(-3)-----(-1)-----(1)----->

Yes, that's correct.

RTCNTC said:
It looks like we must test 5 intervals, correct?

You really only need to test 1 interval, since all roots are of multiplicity 1 (odd), the sign will alternate across all intervals. Because the inequality is weak, we include the end-points of the intervals that are part of the solution. :D
 
Is the correct number line as follows?

<-----(-5)-----(-3)-----(-1)-----(1)----->

I am going to test each interval for algebra practice.

(x - 1)(x + 1)/(x + 3)(x + 5) ≥ 0

When x = -3 & -5, we get undefined. This means we do include -3 and -5 as part of our solution. The same can be said for x = -1 & x = 1.

For (-infinity, -5), let x = -6. Here we get a true statement.

For (-5, -3), let x = -4. Here we get a false statement.

For (-3, -1), let x = -2. Here we get a false statement.

For (-1, 1), let x = 0. Here we get a false statement.

For (1, infinity), let x = 3. Here we get a true statement.

Solution: (-infinity, -5) U (1, infinity)

Correct?
 
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