Finding the discriminant of a quadratic equation from the given information on the roots of a quadratic equation 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?
 A: You at some point got to $\alpha \beta (\alpha-\beta)^2=0.$ But you aren't allowed to cancel the factor $\alpha \beta$ at that point, since it might be zero. In other words you don't really arrive at $\alpha=\beta.$
A: $\alpha^4+\beta^4+2\alpha^2\beta^2=\alpha^4+\beta^4+\alpha\beta^3+\beta\alpha^3$
$2\alpha^2\beta^2=\alpha\beta^3+\beta\alpha^3$
$2\alpha^2\beta^2=\alpha\beta(\alpha^2+\beta^2)$
$\alpha\beta(\alpha^2+\beta^2-2\alpha\beta)=0$
$\alpha\beta(\alpha-\beta)^2=0$
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.
A: Your mistake has been pointed out by the other answers. I just highlight that question should have been - 'which of the following is $\textbf{always}$ correct':
If the roots are equal say equal to $x$, then $\alpha+\beta$, $\alpha^2+ \beta^2$, $\alpha^3+\beta^3$ are $2, 2x^2, 2x^3$, which are in GP. Then $\Delta = 0$ and hence all B, C and D options are correct.
Suppose roots are not equal and that $c=0$. Then the roots are $0$ and $\frac{-b}{a}$. Clearly $\frac{-b}{a}$, $\frac{b^2}{a^2}$ and $\frac{-b^3}{a^3}$ are in GP. Hence option A and C are correct.
A: There is another approach that might be taken that largely avoids the issue of factorization.  While it is tempting upon seeing the expressions $ \ \alpha + \beta \ , \ \alpha^2 + \beta^2 \ , \ \alpha^3 + \beta^3 \  $ to consider an argument based on Newton's identities, that leads to a great deal of writing (and a bit of interpretative complication).  Instead, we might deal with this as a "logic puzzle" by working with the coefficients of the quadratic equation.
The only requirement on the coefficients is that $ \ a \ \neq \ 0 \ \ . $  We can start with the case of $ \ c = 0 \ \   $ only, keeping in mind that $ \ \alpha + \beta \ = \ -\frac{b}{a} \ $ and $ \ \alpha · \beta \ = \ \frac{c}{a} \ \ . $  This tells us that one or both of $ \ \alpha \ $ and $ \ \beta \ $ must equal zero, but the sum of the zeroes requires that one of these be equal to $ \ -\frac{b}{a} \ \ . $  (We also see this by observing that the quadratic equation in this case is $ \ ax^2 + bx = 0 \ \ . $ )  The expressions under discussion then have one term equal to zero, making them fall into the geometric progression $ \ -\frac{b}{a} \ , \ (-\frac{b}{a})^2 \ , \ (-\frac{b}{a})^3 \ \ . $
If we have $ \ b = 0 \ \ $ , then $ \ \alpha + \beta \ = \ 0 \ \Rightarrow \ \alpha \ = \ -\beta \ \ , $ in which case $ \ \alpha^3 + \beta^3 \ = \ 0 \ \ , $ but $ \ \alpha^2 + \beta^2 \ \neq \ 0 \ \   $ generally.  The only exception is to have also $ \ c = 0 \ \ , $ which leads to $ \ \alpha \ = \ -\beta \ = \ 0 \ $ and the trivial geometric progression $ \ 0 \ , \ 0 \ , \ 0 \ \ . $  (The quadratic equation has "collapsed" to $ \ ax^2 \ = \ 0 \ \ . ) \  $    So $ \ b = c = 0 \ $ can be considered a subsidiary case of $ \ c = 0 \ \ . $
If we have for the discriminant $ \ \Delta = 0 \ \ , $ then the quadratic equation has the "double root" $ \ \alpha \ = \ \beta \ = \ -\frac{b}{2a} \ \ . $  We again obtain a geometric progression
$$ \alpha + \beta \ = \ 2 · \alpha \ = \  -\frac{b}{a} \ \ \ , \ \ \ \alpha^2 + \beta^2 \ = \ 2 · \alpha^2 \ = \ 2· \left(-\frac{b}{2a} \right)^2 \ = \ \frac{b^2}{2a^2} \ \ \ , $$
$$ \alpha^3 + \beta^3 \ = \ 2 · \alpha^3 \ = \ 2· \left(-\frac{b}{2a} \right)^3 \ = \ -\frac{b^3}{4a^3} \ \ . $$
It remains to determine whether $ \ \Delta \neq 0 \ $ (two distinct, non-zero roots) can produce a geometric progression.  To save a bit of writing, we will identify the roots by their "components" $ \ \alpha \ = \ B + D \ \ , \ \ \beta \ = \ B - D \ \ . $  The expressions are then
$$ \alpha + \beta \ = \ 2 · B  \ \ \ , \ \ \ \alpha^2 + \beta^2 \ \ = \ \  (B + D)^2 \ + \ (B - D)^2 \ \ = \ \ 2· (B^2 + D^2) \ \ \ , $$
$$ \alpha^3 + \beta^3 \ \ = \ \  (B + D)^3 \ + \ (B - D)^3 \ \ = \ \ 2 B^3 \ + \ 6 B D^2 \ \   . $$
In order for these to be in geometric progression, we require
$$ r \ \ = \ \ \frac{2· (B^2 + D^2)}{2 · B} \ \ = \ \ \frac{2 B^3 \ + \ 6 B D^2}{2· (B^2 + D^2)} $$
$$ \Rightarrow \ \ (B^2 + D^2)^2 \ \ = \ \ B · (B^3 \ + \ 3 B D^2) \ \ \Rightarrow \ \ B^4 \ + \ 2·B^2D^2 \ + \ D^4 \ \ = \ \ B^4 \ + \ 3·B^2D^2$$
$$ \Rightarrow \ \ B^2D^2 \ - \ D^4 \ \ = \ \ D^2 \ · \ (B^2 \ - \ D^2) \ \ = \ \ 0 \ \ . $$
(This is along the lines of the equation you developed.)
This produces no new zeroes:  we have already rejected $ \ D \ = \ \frac{\sqrt{\Delta}}{2a} \ = \ 0 \ \ $ in our assumption, and $ \ B \ = \ \pm D \ $ gives us the zeroes $ \ 0 \ $ and $ \ 2·B \ = \ -\frac{b}{a} \ \   $ (and we will have the associated geometric progression).  This does appear permit the condition $ \ \Delta \ = \ b^2 \ \neq \ 0 \ \ \ , $ but that is equivalent to $ \ b^2 \ = \ b^2 - 4ac \ \Rightarrow \ 4ac \ = \ 0 \ \Rightarrow \ c = 0 \ \ , $ since $ \ a \neq 0 \ \ . $
The proposition can thus be stated as:
If $ \ c = 0 \ \ $ (or $ \ b = 0 \ \   \mathbf{and}  \ c = 0 \     ) \ $ or $ \ \Delta = 0 \ \ $ (or $ \ \Delta = b^2  \ ) \ , $ then $ \ \alpha + \beta \ , \ \alpha^2 + \beta^2 \ , \ \alpha^3 + \beta^3 \  $ are in geometric progression for the real roots $ \ \alpha \ $ and $ \ \beta \ $ of the quadratic equation $ \ ax^2 + bx + c = 0 \ \ . $
This makes choices $ \ \mathbf{(a)} \ \ \text{and} \ \ \mathbf{(b)} \ $ provisionally correct, since they  can be true under   specific conditions . These are really   "gotcha" choices, and shouldn't properly be accepted (or even offered as   choices) on an exam.  Choice $ \ \mathbf{(d)} \ $ is technically correct, but it clearly doesn't "give the whole story".  Choice $ \ \mathbf{(c)} \ $ is the only one that covers the requisite conditions fully.
My feeling is that this is a poorly-constructed set of choices, and perhaps this shouldn't have been posed as a multiple-choice question.  (I am picturing the minor riot that erupted when students contested the instructor's answer.)  I agree with Yathiraj Sharma in that the problem could be "saved" if it instead asks which choice is always correct.
