<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] We have
$$\begin{align}&(a^4+b^4)(c^4+d^4)\\&\ge(a^2c^2+b^2d^2)^2\\&\ge \left(2\sqrt{(a^2c^2)\times (b^2d^2)}\right)^2\\&=4a^2b^2c^2d^2\end{align}$$
Hence, the discriminant 
$$(4abcd)^2-4(a^4+b^4)(c^4+d^4)$$
is non-positive.
Thus, the discriminant has to be $0$, which implies that the real roots are equal.
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