Quadratic Inequality....5

[mathdad]

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?

 

[MarkFL]

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).

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.

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

 

[mathdad]

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?