find x

0=(-2x^2-2)/((x^2-1)^2)

Question

find x

0=(-2x^2-2)/((x^2-1)^2)

agent0smith · Answer

$$\Large 0=\frac{-2x^2-2} {(x^2-1)^2}$$start by multiplying both sides by (x^2-1)^2 

(the denominator can not make a fraction equal to zero, so you can discard it... find where the numerator is zero) $$\Large 0=-2x^2-2$$

agent0smith · Answer

But make sure that you *check* your solutions in this one...

owlcoffee · Answer

Wouldn't it be a little easier to see if she expanded that denominator?

owlcoffee · Answer

I mean, she would be able to simplify the numerator by expanding the denominator, giving a more... Clear answer?

agent0smith · Answer

@Owlcoffee the denominator can never make a fraction equal to zero. It doesn't matter what it is (as long as you check to make sure you arent dividing by zero $$\large 0 = \frac{-2x^2-2} {2^x(x^2-4x^4)^6 (\sin x-4x) \cos^3 x \log (\tan (\cos ln x))} $$this problem would be solved in the same way as the above... all that matters is the numerator.

agent0smith · Answer

$$\Large 0=\frac{-2x^2-2} {(x^2-1)^2} = \frac{-2(x^2+1)} {(x-1)^2(x+1)^2}$$there doesn't appear to be any common factors, and looking for them is more work than solving this

$$\Large 0=-2x^2-2$$and checking solutions.

owlcoffee · Answer

I see, makes sense...
Wish I had known that on my highschool math exam, would've saved me a lot of time.