<s>[INST] roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real 
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.

$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
 [/INST] Solving the equation requires you to compute square roots of the discriminant$$\Delta=(4abcd)^2-4(a^4+b^4)(c^4+d^4).$$If the roots are real, then $\Delta$ has to be non-negative. But$$0\leq (a^2-b^2)=a^4+b^4-2a^2b^2,$$so that$$0\leq 2a^2b^2\leq a^4+b^4$$and similarly$$0\leq 2c^2d^2\leq c^4+d^4.$$Taking the product of these inequalities yields$$4a^2b^2c^2d^2\leq (a^4+b^4)(c^4+d^4),$$so $\Delta \leq 0$. Thus $\Delta=0$, and the roots must both be equal to$$\frac{-4abcd\pm\sqrt{0}}{2(a^4+b^4)}=\frac{-2abcd}{a^4+b^4}.$$
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