Answer to Question #330477 in Algebra for Elley

Question

If we know that the roots of the cubic equation x³ - 61x² -8000 = 0 are in G.P., then find the roots of the equation.

Answer 1

Consider the substitution: $x=z-a$. The equation takes the form: $(z-a)^3-61(z-a)^2-8000=0$. We expand the latter and get: $z^3-3az^2+3a^2z-a^3-61(z^2-2az+z^2)-8000=0.$ Set $a=-\frac{61}{3}$ to delete the term that contains $z^2$. We get: $z^3+pz+q=0$ with $p=-\frac{3721}{3}$, $q=-\frac{669962}{27}$. Compute the following value $\Delta=\frac{q^2}4+\frac{p^3}{27}\approx83253629.63$. One of roots of equation is: $z=(-\frac{q}{2}+\Delta)^{(1/3)}+(-\frac{q}{2}-\Delta)^{(1/3)}\approx42.6813$. The corresponding root of the cubic equation is: $x_0=z-a\approx63.0147$. Consider the function: $f(x)=x^3-61x^2-8000$. The derivative is $f'=3x^2-122x=3x(x-\frac{122}{3})$. From the latter we get: $f'>0$ for $x\in(-\infty,0)\cup(\frac{122}{3},+\infty)$. It means that the function increases on intervals: $(-\infty,0)$ and $(\frac{122}{3},+\infty)$. $f(0)=-8000$. $f(\frac{122}{3})<0$. Thus, the function has only one real root. It is on the interval $(\frac{122}{3},+\infty)$. It means, that it is possible to present the function as: $f(x)=(x-x_0)((x+\alpha)^2+\beta)$with real $\alpha$ and $\beta$. The following relations holds: $2\alpha-x_0=-61$, $-2\alpha x_0+\alpha^2+\beta=0$ We receive: $\alpha=\frac{x_0-61}{2}\approx1.0073$, $\beta=2\alpha x_0-\alpha^2\approx125.9398$. Other two roots have the form: $x_1=-\alpha-i\sqrt{\beta}\approx-1.0073+11.2223\,i,\quad x_2=-\alpha-i\sqrt{\beta}\approx-1.0073-11.2223\,i.$

Thus, roots of equation have the form: $x_0\approx63.0147$, $x_1\approx-1.0073+11.2223\,i$, $x_2\approx-1.0073-11.2223\,i$.