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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?

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  • $\begingroup$ If option D is correct, then C should also be correct and so is B $\endgroup$ Jun 26, 2021 at 4:04
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    $\begingroup$ "will ultimately result in ..." - Show those steps, you lost a factor of $\alpha \beta$ along the way. $\endgroup$
    – dxiv
    Jun 26, 2021 at 4:07

4 Answers 4

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$\alpha^4+\beta^4+2\alpha^2\beta^2=\alpha^4+\beta^4+\alpha\beta^3+\beta\alpha^3$

$2\alpha^2\beta^2=\alpha\beta^3+\beta\alpha^3$

$2\alpha^2\beta^2=\alpha\beta(\alpha^2+\beta^2)$

$\alpha\beta(\alpha^2+\beta^2-2\alpha\beta)=0$

$\alpha\beta(\alpha-\beta)^2=0$

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.

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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 $\alpha=\beta.$

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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.

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    $\begingroup$ Your complete breakdown of the cases is useful. +1. $\endgroup$
    – coffeemath
    Jun 26, 2021 at 4:38
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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.

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