Question

Solve the equation $8x^3-36x^2-18x+18=0$ the root being in A.P

Answer 1

Let the roots be $a-d,a,a+d$

Sum of roots $=-\frac{b^{\prime }}{a^{\prime }}=\frac{36}{8}=3a$

$\Rightarrow a=\frac{12}{8}=\frac{3}{2}$

Product of roots $=\frac{-d}{a}=\frac{-18}{8}=a(a^2-d^2)$

$\Rightarrow \frac{-9}{4}=\frac{27}{8}-\frac{3d}{2}$

$\Rightarrow \frac{3d^2}{2}=\frac{27}{8}+\frac{18}{8}=\frac{45}{8}$

$\Rightarrow d^2=\frac{30}{8}$

$\Rightarrow d=\frac{\sqrt{5}}{2}$

$\Rightarrow$ Roots $=\frac{3}{2}-\frac{\sqrt{5}}{2},\frac{3}{2},\frac{3}{2}+\frac{\sqrt{15}}{2}$.