SOLUTION: solve : {{{x^2(x+8)<0}}}

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Question 247645: solve : x%5E2%28x%2B8%29%3C0
Answer by RicaMath(8) About Me  (Show Source):
You can put this solution on YOUR website!
Here, we see two numbers are being multiplied together in the inequality (x%5E2 and %28x%2B8%29). And we know that the only way to multiply two numbers together and get a number less than 0 is to multiply a positive and a negative number together. (positive * positive = positive; and negative * negative = positive)


Thus, we know that x%5E2%28x%2B8%29%3C0 is true in either of the following situations:


1. x%5E2+%3C+0 and %28x%2B8%29+%3E+0
2. x%5E2+%3E+0 and %28x%2B8%29+%3E+0


If you look at the first statement carefully, you may note that it can never be true. As previously stated, positive * positive = positive; and negative * negative = positive, so there is no way you can take any integer, square it, and get a negative number. This means, we only need to concern ourselves with the second situation, x%5E2+%3E+0 and %28x%2B8%29+%3E+0.
So, let's solve the two inequalities:
x%5E2+%3E+0
x+%3E+0 (get the square root of both sides)


%28x%2B8%29+%3E+0
x+%3E+-8


So, we get x+%3E+-8 as our answer. (because all items > -8 are also > 0, so we use the more restrictive option as our final answer to ensure all possibilities work.)


We can always double check our work. Let's take x+=+-10, as that satisfies our answer condition. Plugging it into our original inequality:
%28-10%29%5E2%28-10%2B8%29+%3C++0
100%28-2%29+%3C++0
-200+%3C++0, which is a true statement.