Question

If $\alpha ,\beta$ are the roots of $x^2+x+1=0$ and $\gamma ,\delta$ are the roots of $x^2+3x+1=0$ then $(\alpha -\gamma )(\beta +\delta )(\alpha +\delta )(\beta -\gamma )=$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

Given: $\alpha ,\beta$ are roots of equation $x^2+x+1=0$, and $\gamma ,\delta$ are the roots of equation $x^2+3x+1=0$

To find the value of $(\alpha -\gamma )(\beta +\delta )(\alpha +\delta )(\beta -\gamma )$

Sol: $\alpha ,\beta$ are roots of equation $x^2+x+1=0$

Hence, $\alpha +\beta =-1,\alpha \beta =\mathrm{1.............}\left(i\right)$

$\gamma ,\delta$ are the roots of equation $x^2+3x+1=0$

Hence $\gamma +\delta =-3,\gamma \delta =\mathrm{1.........}\left(ii\right)$

$(\alpha -\gamma )(\beta +\delta )(\alpha +\delta )(\beta -\gamma )⟹=\left[\alpha \right(\beta +\delta )-\gamma (\beta +\delta \left)\right]\left[\alpha \right(\beta -\gamma )+\delta (\beta -\gamma \left)\right]⟹=(\alpha \beta +\alpha \delta -\gamma \beta -\gamma \delta )(\alpha \beta -\alpha \gamma +\beta \delta -\delta \gamma )$

$⟹=(1+\alpha \delta -\gamma \beta -1)(1-\alpha \gamma +\delta \beta -1)$ [using (i) and (ii)]

$⟹=(\alpha \delta -\gamma \beta )(\delta \beta -\alpha \gamma )⟹=\alpha \delta (\delta \beta -\alpha \gamma )-\gamma \beta (\delta \beta -\alpha \gamma )⟹=\left(\alpha \beta \right)(\delta {)}^2-(\gamma \delta \left)\right(\alpha {)}^2-\left(\gamma \delta \right)(\beta {)}^2+(\alpha \beta \left)\right(\gamma {)}^2$

Substitute $\alpha \beta =1,\gamma \delta =1$, we get

$⟹=({\delta }^2+{\gamma }^2)-({\alpha }^2+{\beta }^2)$

Use formula $(a^2+b^2)=(a+b{)}^2-2ab$

$⟹=(\gamma +\delta {)}^2-2\gamma \delta -[(\alpha +\beta {)}^2-2\alpha \beta ]$

Substitute values of equation (i) and (ii), we get

$⟹=(-3{)}^2-2(1)-[(-1{)}^2-2(1\left)\right]⟹=9-2-1+2⟹=8$

Therefore $(\alpha -\gamma )(\beta +\delta )(\alpha +\delta )(\beta -\gamma )=8$