# 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?

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

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

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

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

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

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

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.

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 $$\alpha=\beta.$$

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.

• Your complete breakdown of the cases is useful. +1. Jun 26, 2021 at 4:38

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.