Quadratic Equations(determine the nature of roots)

Question

$(a^2+b^2)x^2+2(bc+ad)x+(c^2+d^2)=0$ is a quadratic equation, and it's two roots are real, then prove that the roots will be equal.

Sol: From the information given in the question, we can say that discriminant(D) is greater than zero.

$D > 0$

$D = {2(bc+ad)}^2-4(a^2+b^2)(c^2+d^2) = -4(ac-bd)^2$

So here $-4(ac-bd)^2$ cannot be greater than zero.

Can I say that because of $D<0$ so the roots cannot be equal and the assumption that the $D>0$ is also not true?

Answer 1

If the roots are real then $D\ge0$ (not always $D>0$). You have proved that $$D=-4(ac-bd)^2$$ so obviously $D\le0$. Put these together, $D=0$, so the two roots are equal.

Answer 2

By C-S $$(a^2+b^2)(c^2+d^2)=(a^2+b^2)(d^2+c^2)\geq(ad+bc)^2.$$ Thus, $$\frac{\Delta}{4}=(ad+bc)^2-(a^2+b^2)(c^2+d^2)\leq0,$$ which gives that $\Delta=0$ and roots are equal.