I recently came accross an old question that I solved during my school days. Which is
If $\alpha, \beta$ are two real roots of a quadratic equation $ ax^2+bx+c=0 $ and $\alpha+\beta, \alpha^2+\beta^2, \alpha^3+\beta^3$ are in GP,
then which of the following is correct?a) $\Delta\neq0$
b) $b\Delta=0$
c) $c\Delta=0$
d) $\Delta=0$
After seeing the question, I immediately realised that $\alpha+\beta=\frac{-b}{a}$ and $\alpha\beta=\frac{c}{a}$.
And since $\alpha+\beta, \alpha^2+\beta^2, \alpha^3+\beta^3$ are in GP, I got the following equation, I wrote them as $(\alpha^2+\beta^2)^2=(\alpha+\beta)(\alpha^3+\beta^3)$ -> call eqn $i$.
I rewrote the above equation as
$$[(\alpha+\beta)^2-2\alpha\beta]^2=(\alpha+\beta)[(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)]$$
I combined the above two and simplified further which resulted in $ac(b^2-4ac)=0$. And since $a$ can not be zero and $\Delta=b^2-4ac$, I concluded that $c\Delta=0$ and option c is correct.
However I later realised that eqn $i$ can expanded as follows,
$$\alpha^4+\beta^4+2\alpha^2\beta^2=\alpha^4+\beta^4+\alpha\beta^3+\beta\alpha^3$$.
Which will ultimately result in saying that $(\alpha-\beta)^2=0$ and therefore $\alpha=\beta$. If the roots are equal, the discriminant ($\Delta$) has to be zero which means option d is more correct. But most online websites only marked option c as the correct answer.
So which one is the correct answer really? Are they both correct? Or Am I missing something here?
Best Answer
$\alpha^4+\beta^4+2\alpha^2\beta^2=\alpha^4+\beta^4+\alpha\beta^3+\beta\alpha^3$
$2\alpha^2\beta^2=\alpha\beta^3+\beta\alpha^3$
$2\alpha^2\beta^2=\alpha\beta(\alpha^2+\beta^2)$
$\alpha\beta(\alpha^2+\beta^2-2\alpha\beta)=0$
$\alpha\beta(\alpha-\beta)^2=0$
Now this is where you got problem . You cannot cancel $\alpha\beta$ from both sides as we are not sure that they will be non-zero.
Continuing further
$\displaystyle\frac{c}{a}.\frac{\Delta}{a^2}=0$
using the fact that $\alpha\beta=\frac{c}{a}$ and $|\alpha-\beta|=\frac{\sqrt{\Delta}}{|a|}$
therefore , now since we are sure that $a\neq 0$ we can cancel $a^3$ from both sides
$c.\Delta=0$ which is same as you got.