Question

Solve the equation $3x^3-22x^2+48x-32=0$, the roots of which are in harmonic progression.

Answer 1

Given the equation $3x^3-22x^2+48x-32=0$. 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 $p\left(x\right):32x^3-48x^2+22x-3=0$

Sum of the roots, $S_1=a-d+a+a+d=3a=\frac{48}{32}⟹a=\frac{1}{2}$

$\therefore (x-\frac{1}{2})$ is one of the factors of $p\left(x\right)$

$\therefore 32x^3-48x^2+22x-3=(x-\frac{1}{2})(A{x}^2+Bx+C)=A{x}^3+(B-\frac{1}{2}A)x^2+(C-\frac{1}{2}B)x-\frac{1}{2}C$

Equating the coefficients, we get $A=32;B=-32$ and $C=6$

$\therefore p\left(x\right)=(x-\frac{1}{2})(32x^2-32x+6)=(2x-1)(4x–3)(4x-1)$

$\therefore$ the roots of the equation $p\left(x\right)=0$ are $x=\frac{1}{4},\frac{1}{2}$ and $\frac{3}{4}$

$\therefore$ the roots of the original equation are $2,4,$ and $\frac{4}{3}$