I am kind of stuck on the following...

There is a cubic equation like this: $x^3-px^2+qx-r=0$

We are given that the three roots are: $ak^{-1},a,ak$ for some constants $k$ and $a$. Also given (actually one has to deduce this) that $q/p=a$ is one root, and that the product of the other two is $(q/p)^2$ one has to deduce that the roots are in geometric progression... (also note that $r=a^3=(q/p)^3$)

If we let the three roots be: $\alpha,\beta,\gamma$ then $\gamma=\alpha r^2$ and $\beta=\alpha r$. We know that $\alpha \gamma = (q/p)^2$ and that $\alpha \alpha r^2=\alpha^2 r^2=(q/p)^2$. And now I have no clue what to do.


Use that

$$x^3-px^2+qx-r=\left(x-\frac{a}{k}\right)(x-a)(x-ak)=x^3-\left(ak+a+\frac{ a}{k}\right)x^2+\left(a^2k+a^2+\frac{a^2}{k}\right)x-a^3.$$ Then, $\displaystyle\frac{q}{p}=a.$ Moreover, the product of the other two roots is $$ak\cdot \frac{a}{k}=a^2=\left(\frac{q}{p}\right)^2.$$ Finally,

$$ak\cdot a\cdot \frac{a}{k}=a^3=r=\left(\frac{q}{p}\right)^3.$$

  • $\begingroup$ brain freeze.... So we do not show explicitly that one root is $\alpha$, then the other one is $\alpha r$ and the third $\alpha r^2$... rather we use the product of the three roots.. so I had the answer all this time, not realizing it $\endgroup$ – i squared - Keep it Real Jan 16 '16 at 18:25

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