Question #330477

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.

1
Expert's answer
2022-04-25T11:57:46-0400

Consider the substitution: x=zax=z-a. The equation takes the form: (za)361(za)28000=0(z-a)^3-61(z-a)^2-8000=0. We expand the latter and get: z33az2+3a2za361(z22az+z2)8000=0.z^3-3az^2+3a^2z-a^3-61(z^2-2az+z^2)-8000=0. Set a=613a=-\frac{61}{3} to delete the term that contains z2z^2. We get: z3+pz+q=0z^3+pz+q=0 with p=37213p=-\frac{3721}{3}, q=66996227q=-\frac{669962}{27}. Compute the following value Δ=q24+p32783253629.63\Delta=\frac{q^2}4+\frac{p^3}{27}\approx83253629.63. One of roots of equation is: z=(q2+Δ)(1/3)+(q2Δ)(1/3)42.6813z=(-\frac{q}{2}+\Delta)^{(1/3)}+(-\frac{q}{2}-\Delta)^{(1/3)}\approx42.6813. The corresponding root of the cubic equation is: x0=za63.0147x_0=z-a\approx63.0147. Consider the function: f(x)=x361x28000f(x)=x^3-61x^2-8000. The derivative is f=3x2122x=3x(x1223)f'=3x^2-122x=3x(x-\frac{122}{3}). From the latter we get: f>0f'>0 for x(,0)(1223,+)x\in(-\infty,0)\cup(\frac{122}{3},+\infty). It means that the function increases on intervals: (,0)(-\infty,0) and (1223,+)(\frac{122}{3},+\infty). f(0)=8000f(0)=-8000. f(1223)<0f(\frac{122}{3})<0. Thus, the function has only one real root. It is on the interval (1223,+)(\frac{122}{3},+\infty). It means, that it is possible to present the function as: f(x)=(xx0)((x+α)2+β)f(x)=(x-x_0)((x+\alpha)^2+\beta)with real α\alpha and β\beta. The following relations holds: 2αx0=612\alpha-x_0=-61, 2αx0+α2+β=0-2\alpha x_0+\alpha^2+\beta=0 We receive: α=x06121.0073\alpha=\frac{x_0-61}{2}\approx1.0073, β=2αx0α2125.9398\beta=2\alpha x_0-\alpha^2\approx125.9398. Other two roots have the form: x1=αiβ1.0073+11.2223i,x2=αiβ1.007311.2223i.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: x063.0147x_0\approx63.0147, x11.0073+11.2223ix_1\approx-1.0073+11.2223\,i, x21.007311.2223ix_2\approx-1.0073-11.2223\,i.


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