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.
 A: 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.
A: 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$$\frac{-4abcd\pm\sqrt{0}}{2(a^4+b^4)}=\frac{-2abcd}{a^4+b^4}.$$
A: 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.
A: Hint: consider the term $$-4 \left(a^4 c^4+a^4 d^4-4 a^2 b^2 c^2 d^2+b^4 c^4+b^4 d^4\right)$$
A: 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:
$$-\frac{2abcd}{a^4+b^4}.$$
