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.
On
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
$\begin{cases}a\leqslant-6\;\;\lor\;\;a\geqslant2\\b=-2a\quad\qquad\qquad.\end{cases}$
On
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 \ \ . $
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 $2r^2+(a+2)r+2=0$,
$$\Delta=(a+2)^2-16 \ge 0$$