<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] You want to see when the discriminant is non negative, that is,
$$
(4abcd)^2-4(a^4+b^4)(c^4+d^4)\ge0
$$
that can be simplified into
$$
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)\ge0
$$
Consider the left-hand side:
\begin{align}
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)
&=4a^2b^2c^2d^2-a^4c^4-b^4c^4-a^4d^4-b^4d^4\\
&=(-a^4c^4+2a^2b^2c^2d^2-b^4d^4)+(-b^4c^4+2a^2b^2c^2d^2-a^4d^4)\\
&=-(a^2c^2-b^2d^2)^2-(b^2c^2-a^2d^2)^2\\
&=-\bigl((a^2c^2-b^2d^2)^2+(b^2c^2-a^2d^2)^2\bigr)
\end{align}
Since a sum of squares is non negative, the discriminant is non negative if and only if it is zero; this implies the root are coincident, when real.
</s>