Question

Let $a,b$ and $c$ be three distinct real roots of the cubic $x^3+2x^2-4x-4=0$. If the equation $x^3+q{x}^2+rx+s=0$ has roots $\frac{1}{a},\frac{1}{b}$ and $\frac{1}{c}$ , then the value of $(q+r+s)$ is equal to

• A. $\frac{-3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $\frac{-3}{4}$

$x^3+2x^2-4x-4=0\to$ roots a,b,c

$a+b+c=\frac{2}{1}=2,ab+bc+ca=-\frac{(-4)}{1}=4,abc=-4$

$x^3+9x^2+rx+s=0\to$ Roots $\frac{1}{a},\frac{1}{b},\frac{1}{c}$

$q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

$r=-(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$

$s=\frac{1}{abc}$

$q+r+s=\frac{(ab+bc+ca)-(a+b+c)+1}{abc}$

$=\frac{4-2+1}{-4}$

$=\frac{-3}{4}$