Question

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3+3x^2-4x+2=0$ then the equation whose roots are $\frac{1}{\alpha \beta }$, $\frac{1}{\beta \gamma }$,$\frac{1}{\gamma \alpha }$

• A. $4x^3-6x^2+4x+1=0$
• B. $4x^3+6x^2-4x-1=0$
• C. $4x^3+6x^2-4x+1=0$
• D. $4x^3-6x^2-4x-1=0$

Answer 1

The correct option is D $4x^3-6x^2-4x-1=0$

Given: $\alpha ,\beta ,\gamma$ are the roots of $x^3+3x^2-4x+2=0$

To find the equation whose roots are $\frac{1}{\alpha \beta },\frac{1}{\beta \gamma },\frac{1}{\gamma \alpha }$

Sol: We know,

$\left(i\right)\alpha +\beta +\gamma =-3\left(ii\right)\alpha \beta +\beta \gamma +\gamma \alpha =-4\left(iii\right)\alpha \beta \gamma =-2$

Let $\frac{1}{\alpha \beta }=a,\frac{1}{\beta \gamma }=b,\frac{1}{\gamma \alpha }=c$

Then $a,b,c$ are the roots of the desired polynomial. Now,

$\left(i^{\prime }\right)a+b+c=\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=\frac{-3}{-2}=\frac{3}{2}\left(i{i}^{\prime }\right)ab+bc+ca=\frac{1}{\alpha \beta }\times \frac{1}{\beta \gamma }+\frac{1}{\beta \gamma }\times \frac{1}{\gamma \alpha }+\frac{1}{\alpha \beta }\times \frac{1}{\gamma \alpha }=\frac{\alpha \gamma +\alpha \beta +\beta \gamma }{(\alpha \beta \gamma {)}^2}=\frac{-4}{(-2{)}^2}=-1\left(ii{i}^{\prime }\right)abc=\frac{1}{\alpha \beta }\times \frac{1}{\beta \gamma }\times \frac{1}{\alpha \gamma }=\frac{1}{(\alpha \beta \gamma {)}^2}=\frac{1}{(-2{)}^2}=\frac{1}{4}$

From (i'), (ii') and (iii') we get

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0⟹x^3-\frac{3}{2}{x}^2+(-1)x-\frac{1}{4}=0⟹4x^3-6x^2-4x-1=0$

is the required polynomial.