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The given equation is,

$\frac{m}{n}x^2+\frac{n}{m}=1-2x$

What I've tried,

Multiplying the equation by $n$, we get

$mx^2+\frac{n^2}{m}x=n-2nx$

Now what? I am completely confused about what to do. Have I followed the correct steps?

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  • $\begingroup$ group like powers of x and then complete the square $\endgroup$
    – user7530
    Oct 27, 2015 at 16:04
  • $\begingroup$ Things will look nicer if you multiply through by $mn$. $\endgroup$ Oct 27, 2015 at 16:05
  • $\begingroup$ $(mx + n)^2 = mn$ is what you should get if you follow both the above suggestions. I'm sure you can take it up from there. $\endgroup$
    – Shailesh
    Oct 27, 2015 at 16:17

1 Answer 1

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multiply by $\frac{m}{n}$ on both sides, you get

$(\frac{m}{n})^2 x^2+1=(1-2x)\frac{m}{n}$

$(\frac{m}{n})^2 x^2+2x\frac{m}{n} + 1 -\frac{m}{n}=0$

$(\frac{mx}{n})^2+2(\frac{mx}{n}) + (1 -\frac{m}{n})=0$

can you see you got a quadratic equation in $\frac{mx}{n}$?

Do quadratic formula here then divide by $\frac{m}{n}$ and you got what x equals to

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  • $\begingroup$ The title requires the equation to be solved by factorization. $\endgroup$
    – Mick
    Oct 28, 2015 at 15:24
  • $\begingroup$ easy. turn the quadratic equation into a factorization $\endgroup$ Oct 28, 2015 at 15:25
  • $\begingroup$ I don't think that is the aim of this exercise. $\endgroup$
    – Mick
    Oct 28, 2015 at 15:28
  • $\begingroup$ You might be right. I want to know if Abhishekstudent found what he wants, but he hasn't replied yet $\endgroup$ Oct 28, 2015 at 15:33

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