Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+qx+r=0$, then the equation whose roots are $-{\alpha }^{-1},-{\beta }^{-1},-{\gamma }^{-1}$ is:

• A. $r{x}^3+q{x}^2-1=0$
• B. $r{x}^3-q{x}^2-1=0$
• C. $r{x}^3+q{x}^2+1=0$
• D. $r{x}^3-q{x}^2+1=0$

Answer 1

The correct option is B $r{x}^3-q{x}^2-1=0$

where $a,b,c$ are coefficients of $x^3,x^2,x$ respectively and $d$ is the constant

So in equation $x^3+qx+r=0\to \alpha +\beta +\gamma =0\to \alpha \beta +\beta \gamma +\gamma \alpha =q\to \alpha \beta \gamma =-r$

Now equation whose roots are $-{\alpha }^{-1},-{\beta }^{-1},-\gamma -1$

must have sum $=\frac{-1}{\alpha }+\frac{-1}{\beta }+\frac{-1}{\gamma }=\frac{-(\alpha \beta +\beta \gamma +\gamma \alpha )}{\alpha \beta \gamma }=\frac{-q}{r}$

sum of products of two roots at a time =$\frac{-1}{\alpha }\times \frac{-1}{\beta }+\frac{-1}{\beta }\times \frac{-1}{\gamma }+\frac{-1}{\gamma }\times \frac{-1}{\alpha }=\frac{-(\alpha +\beta +\gamma )}{\alpha \beta \gamma }=0$

product of roots = $\frac{-1}{\alpha \beta \gamma }=\frac{1}{r}$

New equation = $x^3-\frac{q}{r}{x}^2-\frac{-1}{r}=0\to r{x}^3-q{x}^2-1=0$

Therfore Answer is $B$