# pre-calculus

solve the inequality algebraically. Express the solution in interval notation. (x-8)^2/x^2-36 > 0

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1. Consider the equation:
(x-8)^2/(x^-36) = 0
(x-8)^2/((x+6)(x-6)) = 0

"critical values are -6,6,8

There are 4 segments on the number line of x which we should look at
1. x < -6
2. x between -6 and 6
3. x between 6 and 8
4. x > 8

I usually take an arbitrary value in each segment, and test it in the original. We don't actually have to work out the value, just consider the signs.
In this case, since the top is squared, we only have to look at the denominator's sign.

1. let x = -10 ----+/+ > 0 , good
2. let x = 0, --- +/- < 0 , no good
3. let x = 7, --- +/+> 0 , good
4. let x = 10 --- +/+ > 0 good

so x < -6 OR x > 6, x is a real number.

(The above is a rather general solution. In this case we could have just looked at the denominator.
clearly we cannot have it become negative. It was easy to see that would happen between -6 and 6 for values of x)

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2. i don't get it so whick would be the solution or answer

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3. "so x < -6 OR x > 6, x is a real number"
As Reiny explained, there are two segments to the answer, hence the "OR".

In interval notation, it would be:
(-∞,-6)∪(6,∞).
The (round) parentheses mean that the boundary values are excluded from the interval.

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