# Find the relation between $m$ and $n$ such that the following equation has four roots. [closed]

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.

• 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? – Ross Millikan Mar 17 '19 at 3:30
• Well, you add $-\dfrac{2mx^2}{m + x}$ to both sides. – Lê Thành Đạt Mar 17 '19 at 3:36

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.

Hint:

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

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

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

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