Question

Solve the equation $40x^4-22x^3-21x^2+2x+1=0$, the roots of which are in harmonic progression.

Answer 1

Given the equation $40x^4-22x^3-21x^2+2x+1=0$. Let the roots of the equation be $\frac{1}{a-3d},\frac{1}{a-d},\frac{1}{a+d}$ and $\frac{1}{a+3d}(\because \text{the roots are in harmonic progression})$

Transform the equation by inputting $x=\frac{1}{x}$ which has the roots $(a-3d),(a-d),(a+d)$ and $(a+3d)$

New equation $p\left(x\right):x^4+2x^3-21x^2-22x+40=0$

Sum of the roots, $S_1=a-3d+a-d+a+d+a+3d=4a=-2⟹a=\frac{-1}{2}$

Sum of the roots taken 2 at a time, $S_2=6a^2–10d^2=-21⟹d=\frac{3}{2}$

$\therefore$ The roots of the equation are: $a-3d=-5;a-d=-2;a+d=1$ and $a+3d=4$

$\therefore$ the roots of the original equation are $\frac{-1}{5},\frac{-1}{2},1$ and $\frac{1}{4}$