Question

Let $a, b$ and $c$ be three distinct real roots of the cubic $x^3+2 x^2-4 x-4=0$. If the equation $x^3+q x^2+r x+s=0$ 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

Answer: (C)

We have
$x^3+2 x^2-4 x-4=0$ has roots, $a, b$ and $c$.....(1)
On replacing $x$ by $\frac{1}{x}$ in equation (1), we get
$4 x^3+4 x^2-2 x-1=0 \Rightarrow x^3+x^2-\frac{1}{2} x-\frac{1}{4}=0$,
which has roots $\frac{1}{a}, \frac{1}{b}$ and $\frac{1}{c}$. ....(2)
$\therefore$ On comparing equation (2) with $x ^3+ qx ^2+ rx + s =0$, we get
$q=1, r=\frac{-1}{2}, s=\frac{-1}{4}$
Hence $(q+r+s)=\frac{1}{4}$