Question

Solve the equation 8x^3-36 x^2-18x+81=0 the roots being in A.P.

Answer 1

$Given \: equation ,$

$8x^3-36 x^2-18x+81=0$

$Let \: the \: roots \: be \: (a-d) , a , (a+d)$

$i )Sum \:of \: roots = \frac{- Coefficient \:of \:x^{2}}{Coefficient \:of \: x^{3}}$

$\implies a-d+a+a+d = \frac{-(-36)}{8}$

$\implies 3a = \frac{9}{2}$

$\implies a = \frac{9}{3\times 2}$

$\implies a = \frac{3}{2} \: --(1)$

$ii ) Product \:of \: roots = \frac{- Constant \:term}{Coefficient \:of \: x^{3}}$

$\implies (a-d)\times a\times (a+d) = \frac{-81}{8}$

$\implies a(a^{2} - d^{2}) = \frac{-81}{8}$

$\implies \frac{3}{2} [\Big(\frac{3}{2}\Big)^{2} -d^{2}] = \frac{-81}{8}$

$\implies \frac{9}{4} - d^{2} = \frac{-27}{4} \times$

$\implies - d^{2} = \frac{-27}{4} - \frac{9}{4}$

$\implies - d^{2} = \frac{-36}{4}$

$\implies d^{2} = 9$

$\implies d = \pm 3$

Therefore.,

$\red{ Required \: roots \:are }$

$If \: a = \frac{3}{2} \: and \: d = 3$

$i )a - d$

$= \frac{3}{2} - 3$

$= \frac{3-6}{2}$

$= \frac{-3}{2}$

$ii ) a = \frac{3}{2}$

$i )a - d$

$= \frac{3}{2} + 3$

$= \frac{3+6}{2}$

$= \frac{9}{2}$

$\green { Roots \: \frac{-3}{2} , \frac{3}{2}, \frac{9}{2}}$

Or

$\red{ Required \: roots \:are }$

$If \: a = \frac{3}{2} \: and \: d = - 3$

$\green { Roots \: \frac{9}{2} , \frac{3}{2}, \frac{-3}{2}}$

•••♪

Answer 2

let the roots be a-d,a,a+d

sum of roots a-d+a+a+d=36/8

3a=36/8

a=3/2

product of roots=a(a^2-b^2) =-81/8

3/2(3/2^2-d^2)=-81/8

9/4-d^2=-27/4

-d^2=-36/4

d^2=9

d=+3 or-3

a=3/2,d=3

a-d=-3/2

2.a=3/2,d=-3

a-d=9/2

roots are -3/2,3/2,9/2