<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] The discriminant is non-positive because of the geometric-quadratic means inequality. Indeed the reduced discriminant is 
$$\Delta'=4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)$$
Sert $A=a^2, B=b^2, \&c.$ We have to prove $\;4ABCD\ge (A^2+B^2)(C^2+D^2)$
Now by the GQM inequality, we have
\begin{alignat*}{2}\sqrt{AB}&\le\sqrt\frac{A^2+B^2}2, &\quad\sqrt{CD}&\le\sqrt\frac{C^2+D^2}2\\
\iff 2AB&\le A^2+B^2,&2CD & \le C^2+D^2
 \end{alignat*}
so multiplying both sides of the inequalities, we get
$$4ABCD\le(A^2+B^2)( C^2+D^2) $$
Thus we have real roots if and only if $4ABCD=(A^2+B^2)( C^2+D^2) $, i. e. if there is a double root, which is necessarily equal to the half sum of the roots by Vieta's formulae:
$$-\frac{2abcd}{a^4+b^4}.$$
</s>