Question

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

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

Answer 1

The correct option is A $x^3+2x^2+3x-1=0$

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

To find the equation whose roots are $\beta +\gamma ,\gamma +\alpha ,\alpha +\beta$

Sol: We know,

$\left(i\right)\alpha +\beta +\gamma =-1\left(ii\right)\alpha \beta +\beta \gamma +\gamma \alpha =2\left(iii\right)\alpha \beta \gamma =-3$

Let $\beta +\gamma =a,\gamma +\alpha =b,\alpha +\beta =c$

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

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

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

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0⟹x^3-(-2)x^2+\left(3\right)x-\left(1\right)=0⟹x^3+2x^2+3x-1=0$

is the required polynomial.