# STUCK ON THIS QUESTION, HELP NEEDED!!!!!!!!?

Im stuck on these questions help is needed plz

### 1 Answer

- anonymousLv 71 month ago
(a) First you need to factor the quadratic on the left side.

-x² - 10x - 9 ≥ 0

(-x - 9)(x + 1) ≥ 0

The values of x that make the above equal to zero are x = -9 and x = -1. This splits the number line into three regions:

x < -9

-9 < x < -1

x > -1

For the sign chart, choose a value in x in each of the above intervals and work out the sign of each factor. For example, for the interval -9 < x < -1, I chose x = -2 and worked out that (-x - 9) = (-2 - 9) = -11, so that factor is negative in that interval. Similarly, (x + 1) = (-2 + 1) = -1, so that factor is also negative in that interval. The product of the two factors is a negative number times a negative number, so the product is positive.

…..……..…. (-x - 9) ….... (x + 1) …….. (-x - 9)(x + 1)

x < -9 ……..…. + …………. - ……………….. -

-9 < x < -1 …… - …………. - ……………….. +

x > -1 ……….... - …………. + ………………. -

The left side of the inequality is greater than zero on - 9 < x < -1 and equal to zero when x = -9 and -1. Taken together, the inequality -x² - 10x - 9 ≥ 0 is true on -9 ≤ x ≤ -1 .

Graph:

https://www.desmos.com/calculator/q0hpv7qtks

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(b) -6x(x + 1)²(3 - x) < 0

Pretend that it is an equality for the moment. The values of x that would make it equal to zero are x = 0, x = -1, and x = 3 . This splits the number line into four regions, which I'll list down the left side of the sign chart:

…………………. -6x .… (x + 1)² …. (3 - x) ….. -6x(x + 1)²(3 - x)

x < -1 …………… + ………+ ………… + …………… +

-1 < x < 0 ………. + ……... + ……….... + …………… +

0 < x < 3 ……….. - …….... + ……….... + …………… -

x > 3 ……………. - …….... + …………. - ………….... +

The left side of the original inequality is only less than zero on one of those regions.

The inequality is true on 0 < x < 3.

Graph (coordinate axes are on different scales) :