# Find all $a,b$ for which the polynomial has real roots and are in geometric progression.

Find all $$a, b$$ such that the roots of $$x^3 + ax^2 + bx − 8 = 0$$ are real and in a geometric progression.

I did deduce the answer till $$a=\dfrac{-b}{2}$$.

Using the Vieta's relations I deduced that if $$\alpha,\beta$$ and $$\gamma$$ are the roots of the equation (in the same order) then $$\beta^2 = \alpha \gamma$$, which implies $$\beta^3 =8$$ or $$\beta=2$$, and using the other two relations, I found that $$a=\dfrac{-b}{2},$$ but got stuck while writing the values of $$a$$ for which the equation satisfies.

• The solution is $\,a\leqslant-6\lor a\geqslant2\,$ and $b=-2a$. Aug 20, 2021 at 19:49

Let $$\dfrac{2}{r}$$, $$2$$ and $$2r$$ be the roots.

Since you have $$b=-2a$$,

$$8r^3+4ar^2-4ar-8=0$$ $$(r-1)[2r^2+(a+2)r+2]=0$$

Now check the discriminant in $$2r^2+(a+2)r+2=0$$,

$$\Delta=(a+2)^2-16 \ge 0$$

• Thanks for the idea Aug 25, 2021 at 15:12

Let $$\,\alpha,\beta,\gamma\,$$ be roots of the equation, then we have $$\,\beta^2=\alpha\gamma\,.$$

Also, from Vieta’s relation, we get that

$$\alpha\beta\gamma=8\implies\beta^3=8\implies\beta=2\;.$$

Now $$\,\alpha+\gamma=-2-a\,$$ and $$\,2\alpha+2\gamma+4=b\,,\,$$ hence ,

$$\begin{cases} \alpha+\gamma=-2-a\\ \alpha+\gamma=-2+\dfrac b2 \end{cases}\quad\implies b=-2a\,.$$

Since $$\,\alpha\,$$ and $$\,\gamma\,$$ are both positive or both negative, it results that $$\,|\alpha+\gamma|=|\alpha|+|\gamma|\,$$ and $$\,|\alpha||\gamma|=\alpha\gamma\,.$$

Moreover, by applying the AM-GM inequality, it follows that

$$|-2-a|=|\alpha+\gamma|=|\alpha|+|\gamma|\geqslant2\sqrt{|\alpha||\gamma|}=2\sqrt{\alpha\gamma}=4\;\;,$$

$$|2+a|\geqslant4\;\;,$$

$$2+a\geqslant4\;\;\;\lor\;\;\;2+a\leqslant-4\;\;,$$

$$a\geqslant2\;\;\lor\;\;a\leqslant-6\;.$$

Consequently the solutions are

$$\begin{cases}a\leqslant-6\;\;\lor\;\;a\geqslant2\\b=-2a\quad\qquad\qquad.\end{cases}$$

We can also work with the zeroes $$\ r \ , \ rs \ , \ rs^2 \ \ ,$$ with $$\ s \neq 1 \$$ being the ratio between "successive" values. Since the cubic polynomial is monic, by direct multiplication of factors or the Viete relations, we obtain $$\ a \ = \ -r·(1 + s + s^2) \ \ , \ \ b \ = \ r^2·s \ + \ r^2·s^2 \ + \ r^2·s^3 \ = \ r^2s · (1 + s + s^2) \ \ ,$$ $$c \ = \ -r^3·s^3 \ = \ -8 \ \ .$$ Consequently, $$\ rs \ = \ 2 \ \$$ and $$\ b \ = \ rs · r·(1 + s + s^2) \ = \ 2 · (-a) \ \ .$$

Thus, we may write the polynomial as $$\ x^3 + ax^2 - 2ax - 8 \ \ .$$ A check by synthetic or polynomial division reveals that this has the factorization $$\ (x - 2)·(x^2 \ + \ [a + 2]·x \ + \ 4) \ \ .$$ So $$\ x = 2 \$$ is always a real zero of the polynomial; we already know that the $$\ y-$$intercept is $$\ (0 \ , \ -8) \ \ .$$ Since we want all three zeroes to be real, the graph of the polynomial must have two turning points and we need to find at least the "threshold" values of $$\ a \$$ which permit all the zeroes to be real. Without calculus, we can use the discriminant of the quadratic factors to find that one of the turning points is a "touching" $$\ x-$$intercept when $$\ (a + 2)^2 \ - \ 4·1·4 \ = \ a^2 + 4a - 12 \ = \ (a - 2)·(a + 6) \ = \ 0 \ \$$ (as Ng Chung Tak also shows by a related approach).

[By using calculus, we find the turning points to be located at $$\frac{d}{dx} \ [ \ x^3 + ax^2 - 2ax - 8 \ ] \ = \ 3x^2 + 2ax - 2a \ = \ 0 \ \Rightarrow \ \ x \ \ = \ \ -\frac{a}{3} \ \pm \ \frac{\sqrt{a^2 + 6a}}{3} \ \ .$$ The "upper" turning point, or relative maximum, of the curve occurs where $$\ \frac{d^2}{dx^2} \ [ \ x^3 + ax^2 - 2ax - 8 \ ] \ = \ 6x + 2a \ < \ 0 \ \ ,$$ which is at $$\ x \ = \ -\frac{a}{3} \ - \ \frac{\sqrt{a^2 + 6a}}{3} \ \ . \ ]$$

These present two different situations. For $$\ a \ = \ -6 \ \ ,$$ the cubic polynomial is $$\ x^3 - 6x^2 + 12x - 8 \ = \ (x - 2)^3 \ \ ,$$ which has a "triple zero" at $$\ x \ = \ 2 \ \ ; \$$ [Here, $$\ 6x + 2a \ = \ 0 \$$ and the "turning point" becomes an inflection point.] Using values $$\ a \ < \ -6 \ \$$ "breaks this up" into the three real zeroes $$\ r = \frac{2}{s} \ , \ rs = 2 \ \ ,$$ and $$\ rs^2 = 2s \ \ .$$ Comparing the appropriate factors with the quadratic factor above, we have $$\left(x - \frac{2}{s} \right) · (x - 2s) \ \ = \ \ x^2 \ + \ [a + 2]·x \ + \ 4 \ \ \Rightarrow \ \ \Rightarrow \ \ -\left(\frac{2}{s} + 2s \right) \ \ = \ \ a \ + \ 2$$ $$\Rightarrow \ \ 2s^2 \ + \ (a+2)·s \ + \ 2 \ \ = \ \ 0 \ \ .$$ This quadratic polynomial has the significant feature that it is palindromic, that is, it has the form $$\ Ax^2 + Bx + A \ \ ;$$ its zeroes are then related as $$\ r \$$ and $$\ \frac{1}{r} \ .$$ What that means for this most recent quadratic equation is that the two possible ratios of geometric progression are reciprocals of one another; in other words, the ratio $$\ s \$$ among the zeroes "read left to right" is the reciprocal ratio for the same zeroes "read right to left". We find that for $$\ a \ < \ -6 \ \ ,$$ these ratios are $$s \ \ = \ \ \frac{-(a + 2) \ + \ \sqrt{a^2 + 4a -12}}{4} \ \ ,$$ $$\frac{1}{s} \ \ = \ \ \frac{-(a + 2) \ - \ \sqrt{a^2 + 4a - 12}}{4} \ \ = \ \ \frac{4}{-(a + 2) \ + \ \sqrt{a^2 + 4a -12}} \ \ .$$

As an illustration, for $$\ a \ = \ -7 \ \ ,$$ the three real zeroes of $$\ x^3 - 7x^2 + 14x - 8 \ \$$ are $$\ 2 \$$ and $$\ -\frac{-7 + 2}{2} \pm \ \frac{\sqrt{[-7]^2 + 4·[-7] - 12}}{2} \ = \ \frac52 \pm \frac32 \ \$$ or $$\ 1 \ , \ 2 \ , \ 4 \ \ .$$ The geometric ratios are found from $$\ -\frac{-7 + 2}{4} \pm \ \frac{\sqrt{[-7]^2 + 4·[-7] - 12}}{4} \ = \ \frac54 \pm \frac34 \ = \ 2 \ , \ \frac12 \ \ ,$$ which we see are the ratios of the geometric progression among the zeroes "read in opposite directions".

For $$\ a \ = \ 2 \ ,$$ the curve for $$\ x^3 + 2x^2 - 4x - 8 \ = \ (x - 2)·(x^2 \ + \ 4·x \ + \ 4) \ = \ (x - 2)·(x + 2)^2 \ \ ,$$ the "upper" turning point at $$x \ = \ -\frac{2}{3} \ - \ \frac{\sqrt{2^2 + 6·2}}{3} \ = \ -2 \ \$$ is a "double zero" of the polynomial. With $$\ a \ > \ 2 \ \ ,$$ this "splits into" two zeroes, with the third still at $$\ x \ = \ +2 \ \ ;$$ the only way $$\ \frac{2}{s} \ , \ 2 \ , \ 2s \$$ could make sense as a geometric progression is if the ratio is negative .

Indeed, for $$\ a \ = \ 3 \ \ ,$$ the polynomial $$\ x^3 + 3x^2 - 6x - 8 \ \$$ has the two negative real zeroes $$\ x^3 - 7x^2 + 14x - 8 \ \$$ are $$\ -\frac{3 + 2}{2} \pm \ \frac{\sqrt{3^2 + 4·3 - 12}}{2} \ = \ -\frac52 \pm \frac32 \ \$$ and the geometric ratios are $$\ -\frac{3 + 2}{4} \pm \ \frac{\sqrt{3^2 + 4·3 - 12}}{4} \ = \ -\frac54 \pm \frac34 \ = \ -2 \ , \ -\frac12 \ \ .$$ Thus, we have the progression of real zeroes $$\ -1 \ , \ 2 \ , \ -4 \ \$$ with the ratios in reciprocal relation corresponding to "reading" the sequence in opposite directions.

So using $$\ a \ < \ -6 \ \ [ \ b \ = \ -2a \ ] \ \$$ yields cubic polynomials with the three real zeroes in a geometric progression with a positive ratio (the second ratio is just another way of interpreting the sequence), while $$\ a \ > \ 2 \ \ [ \ b \ = \ -2a \ ] \ \$$ corresponds to having a progression of three real zeroes with negative ratio (again, the second ratio being auxiliary). The "degenerate" cases are $$\ a \ = \ -6 \ \ ,$$ which has the zeroes $$\ 2 \ , \ 2 \ , \ 2 \$$ with ratio $$\ s \ = \ 1 \ \$$ and $$\ a = +2 \ \$$ with the zeroes being $$\ -2 \ , \ 2 \ , \ -2 \$$ with $$\ s \ = \ -1 \ \ .$$