Question

If $\alpha$, $\beta$, $\gamma$ are the roots of the equation $x^3+qx+r=0$, then the equation whose roots are $\frac{\beta +\gamma }{{\alpha }^2},\frac{\gamma +\alpha }{{\beta }^2},\frac{\alpha +\beta }{{\gamma }^2}$, is

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

Answer 1

Answer: (B)

$r{x}^3-q{x}^2-1=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+qx+r=0$
Then
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\gamma \alpha =q{s}_3=\alpha \beta \gamma =-r$
Now for roots $\frac{\beta +\gamma }{{\alpha }^2},\frac{\gamma +\alpha }{{\beta }^2},\frac{\alpha +\beta }{{\gamma }^2}$
$S_1=\frac{\beta +\gamma }{{\alpha }^2}+\frac{\gamma +\alpha }{{\beta }^2}+\frac{\alpha +\beta }{{\gamma }^2}=\frac{\beta +\gamma +\alpha -\alpha }{{\alpha }^2}+\frac{\gamma +\alpha +\beta -\beta }{{\beta }^2}+\frac{\alpha +\beta +\gamma -\gamma }{{\gamma }^2}=-\frac{1}{\alpha }-\frac{1}{\beta }-\frac{1}{\gamma }=-\left[\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right]=\frac{q}{r}{S}_2=\frac{(\beta +\gamma )(\gamma +\alpha )}{{\alpha }^2{\beta }^2}+\frac{(\gamma +\alpha )(\alpha +\beta )}{{\beta }^2{\gamma }^2}+\frac{(\alpha +\beta )(\beta +\gamma )}{{\alpha }^2{\gamma }^2}$

$=\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=0S_3=\frac{(\beta +\gamma )(\gamma +\alpha )(\alpha +\beta )}{{\left(\alpha \beta \gamma \right)}^2}=-\frac{1}{\alpha \beta \gamma }=\frac{1}{r}$
Hence, the required equation is
$x^3-S_1{x}^2+S_{2}x-S_3=0$

$\Rightarrow x^3-\frac{q}{r}{x}^2+x\left(0\right)-\frac{1}{r}=0$

$\Rightarrow r{x}^3-q{x}^2-1=0$