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Find the relation between $m$ and $n$ such that the following equation has four roots with $m > 0$. $$x^2 + \left(\dfrac{mx}{m + x}\right)^2 = n$$

Well, I know what the answer is. I just want to know the complete answer.

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  • $\begingroup$ If you know the answer you should supply it. It can help others check their work, if nothing else. Then, what have you tried? Where are you stuck? $\endgroup$ – Ross Millikan Mar 17 '19 at 3:30
  • $\begingroup$ Well, you add $-\dfrac{2mx^2}{m + x}$ to both sides. $\endgroup$ – Lê Thành Đạt Mar 17 '19 at 3:36
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It's $$x^2-\frac{2mx^2}{m+x}+\frac{m^2x^2}{(m+x)^2}+\frac{2mx^2}{m+x}=n$$ or $$\left(x-\frac{mx}{m+x}\right)^2+\frac{2mx^2}{m+x}=n$$ or $$\left(\frac{x^2}{m+x}\right)^2+\frac{2mx^2}{m+x}=n.$$ Let $\frac{x^2}{m+x}=t$.

Thus, $$t^2+2mt-n=0,$$ which gives the first condition: $$m^2+n\geq0.$$ We have: $$\frac{x^2}{m+x}=-m+\sqrt{m^2+n}$$ or $$\frac{x^2}{m+x}=-m-\sqrt{m^2+n}.$$ Now, just write $\Delta\geq0$ for these quadratic equations.

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Hint:

If $a=\dfrac{mx}{m+x}, b=x$

as $\dfrac{b-a}{ab}=\dfrac1a-\dfrac1b=\dfrac1m, ab=m(b-a)$

Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$,

$$n=a^2+b^2=(b-a)^2+2ab=(b-a)^2+2m(b-a)$$

which is on re-arrangement, a quadratic eqaution in $b-a$

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