Question

Can you walk me through this problem?

Answer 1

\[2-\frac{ 4 }{ x^2}=-\frac{ 2 }{ x }\]

Answer 2

First multiply by \(x^2\) to get rid of those ugly denominators.
\[x^2\left(2 - \frac{4}{x^2}\right) = x^2\left(-\frac{2}{x}\right)\]\[2x^2 - 4 = -2x\]
Now it's just a quadratic equation.
\[2x^2 + 2x - 4 = 0\]

Answer 3

Divide by 2 to clean it up.
\[x^2 + x - 2 = 0\]
That factors easily.
\[(x - 1)(x + 2) = 0\]
So there are your solutions. :D

Do you understand it now?

Answer 4

Thank you, that helped so much! Just a quick question, I always have trouble with the first step, which in this case was multiplying x^2 by the both sides of the equation. I don't quite understand how to really do that.

Answer 5

Do you mean you don't understand "why \(x^2\)" or how to actually do that computation?

Answer 6

How to actually do the computation. I know it sounds stupid, but that is what I have trouble with.

Answer 7

Ok. Let's do it in detail, then.
\[x^2\left(2 - \frac{4}{x^2}\right) = x^2\left(-\frac{2}{x}\right)\]
Due to the distributive property, \(a(b + c) = ab + ac\) we get:
\[\left(2x^2 - \frac{4x^2}{x^2}\right) = -\frac{2x^2}{x}\]

Answer 8

Write out the squares as a product.
\[2x^2 - \frac{4xx}{xx} = -\frac{2xx}{x}\]
Since \(\frac{a}{a} = 1\) and \(1 \times a = a\) we can just cross out some x's.
\[2x^2 - \frac{4\cancel{xx}}{\cancel{xx}} = -\frac{2\cancel{x}x}{\cancel{x}}\]

Answer 9

That leaves us with
\[2x^2 - \frac{4}{1} = - \frac{2x}{1}\]
Anything over 1 is just equal to itself so that gives us our quadratic.

Answer 10

Hopefully that helped and was enough detail. :D