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.
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Expert's answer
2022-04-25T11:57:46-0400
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: z3−3az2+3a2z−a3−61(z2−2az+z2)−8000=0. Set a=−361 to delete the term that contains z2. We get: z3+pz+q=0 with p=−33721, q=−27669962. Compute the following value Δ=4q2+27p3≈83253629.63. One of roots of equation is: z=(−2q+Δ)(1/3)+(−2q−Δ)(1/3)≈42.6813. The corresponding root of the cubic equation is: x0=z−a≈63.0147. Consider the function: f(x)=x3−61x2−8000. The derivative is f′=3x2−122x=3x(x−3122). From the latter we get: f′>0 for x∈(−∞,0)∪(3122,+∞). It means that the function increases on intervals: (−∞,0) and (3122,+∞). f(0)=−8000. f(3122)<0. Thus, the function has only one real root. It is on the interval (3122,+∞). It means, that it is possible to present the function as: f(x)=(x−x0)((x+α)2+β)with real α and β. The following relations holds: 2α−x0=−61, −2αx0+α2+β=0 We receive: α=2x0−61≈1.0073, β=2αx0−α2≈125.9398. Other two roots have the form: x1=−α−iβ≈−1.0073+11.2223i,x2=−α−iβ≈−1.0073−11.2223i.
Thus, roots of equation have the form: x0≈63.0147, x1≈−1.0073+11.2223i, x2≈−1.0073−11.2223i.
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