Question

The roots of the equation $x^3-11x^2+36x-36=0$ are in harmonic progression; find them.

Answer 1

Given the equation $x^3-11x^2+36x-36=0$. Let the roots of the equation be $\frac{1}{a-d},\frac{1}{a}$ and $\frac{1}{a+d}(\because \text{the roots are in harmonic progression})$

Transform the equation by inputting $x=\frac{1}{x}$ which has the roots $(a-d),a$ and $(a+d)$

New equation $p\left(x\right):36x^3-36x^2+11x-1=0$

Sum of the roots, $S_1=a-d+a+a+d=3a=\frac{36}{36}⟹a=\frac{1}{3}$

$\therefore (x-\frac{1}{3})$ is one of the factors of $p\left(x\right)$

$\therefore 36x^3-36x^2+11x-1=(x-\frac{1}{3})(A{x}^2+Bx+C)=A{x}^3+(B-\frac{1}{3}A)x^2+(C-\frac{1}{3}B)x-\frac{1}{3}C$

Equating the coefficients, we get $A=36;B=-24$ and $C=3$

$\therefore p\left(x\right)=(x-\frac{1}{3})(36x^2-24x+3)=(3x-1)(2x–1)(6x-1)$

$\therefore$ the roots of the equation $p\left(x\right)=0$ are $x=\frac{1}{6},\frac{1}{3}$ and $\frac{1}{2}$

$\therefore$ the roots of the original equation are $2,3,$ and $6$