Question

Three real roots of the equation $2x^3-19x^2+57x+2\lambda =0$ are consecutive term of geometrical progression, then the value $\lambda$ is

• A. $54$
• B. $-54$
• C. $-27$
• D. $-108$

Answer 1

The correct option is C $-27$

Let the consecutive terms of G.P be $\frac{A}{r},A,Ar$
now we know that the sum of the roots of a cubic equation taken one at a time =$\frac{-b}{a}$
hence $\frac{A}{r}+A+Ar=\frac{19}{2}$
and sum of the roots taken two at a time =$\frac{c}{a}$
hence $A^2+\frac{A^2}{r}+A^{2}r=\frac{57}{2}$
dividing the first two equations we get $\frac{A^2(r+\frac{1}{r}+1)}{A(r+\frac{1}{r}+1)}=\frac{57}{19}=3\Rightarrow A=3$
and sum of roots taken 3 at a time =$\frac{-d}{a}$
hence $\frac{A}{r}\times A\times Ar=A^3=-\lambda$
$\Rightarrow \lambda =-27$