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How can I solve this equality:

$ \dfrac{4}{x^2} + \dfrac{2}{x^2-9} = 1 $

I know how to solve it when the RHS would have been $0$ (cross-multiplication).

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Multiply both sides by $x^2\cdot (x^2-9)$ to get $$4(x^2-9)+2x^2=1\cdot x^2\cdot (x^2-9)$$ Simplify both sides and set the right-hand side equal to 0: $$x^4-15x^2+36=0$$ Let $w=x^2$ so that this equation reduces to $$w^2-15w+36=0$$ This is a quadratic you can solve in $w$, then back-substitute to x to solve for $x$.

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Hint: take a common denominator, cross multiply. If you are still not able to come to the solution I will help you then.

Here are the first few steps:

$$\frac{4(x^2-9)+2x^2}{x^4-9x^2}\rightarrow 6(x^2-6)=x^4-9x^2 \rightarrow x^4-16x^2+36=0$$

Now use the substitution $y=x^2$, and then you must only solve something familiar!

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You can still cross multiply and add the fractions to get $4(x^2-9)+2x^2=x^2(x^2-9)$. Though this looks like a quartic, let $y=x^2$ and you have a quadratic in $y$.

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