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 $\left(\alpha -\gamma \right)\left(\beta +\delta \right)\left(\alpha +\delta \right)\left(\beta -\gamma \right)$ is equal to:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D

$8$

Step 1: Find the sum and product of the roots of the equation $x^2+x+1=0$.

$\alpha$ and $\beta$ are the roots of the equation $x^2+x+1=0$. The sum of the roots is:

$\begin{array}{rcl}\alpha +\beta & =& \frac{-\text{coefficient of}x}{\text{Coefficient of}{x}^2}\\ \alpha +\beta & =& \frac{-1}{1}\\ \alpha +\beta & =& -1\end{array}$

The product of the roots is:

$\begin{array}{rcl}\alpha \beta & =& \frac{\text{Constant term}}{\text{Coefficient of}x}\\ \alpha \beta & =& \frac{1}{1}\\ \alpha \beta & =& 1\end{array}$

Step 2: Find the sum and product of the roots of the equation $x^2+3x+1=0$.

$\gamma$ and $\delta$ are the roots of the equation $x^2+3x+1=0$. The sum of the roots is:

$\begin{array}{rcl}\gamma +\delta & =& \frac{-\text{coefficient of}x}{\text{Coefficient of}{x}^2}\\ \gamma +\delta & =& \frac{-3}{1}\\ \gamma +\delta & =& -3\end{array}$

The product of the roots is:

$\begin{array}{rcl}\gamma \delta & =& \frac{\text{Constant term}}{\text{Coefficient of}x}\\ \gamma \delta & =& \frac{1}{1}\\ \gamma \delta & =& 1\end{array}$

Step 3: Find the value of $\left(\alpha -\gamma \right)\left(\beta +\delta \right)\left(\alpha +\delta \right)\left(\beta -\gamma \right)$.

Moreover,

$$\begin{gathered} {\left(\gamma +\delta \right)}^2={\gamma }^2+{\delta }^2+2\gamma \delta \\ {(-3)}^2={\gamma }^2+{\delta }^2+2\left(1\right) \\ 9={\gamma }^2+{\delta }^2+2 \\ {\gamma }^2+{\delta }^2=7 \end{gathered}$$

Therefore,

$\begin{array}{rcl}\left(\alpha -\gamma \right)\left(\beta +\delta \right)\left(\alpha +\delta \right)\left(\beta -\gamma \right)& =& (\alpha -\gamma )(\beta -\gamma )(\beta +\delta )(\alpha +\delta )\\ & =& (\alpha \beta -\alpha \gamma -\beta \gamma +{\gamma }^2)(\alpha \beta +\alpha \delta +\beta \delta +{\delta }^2)\\ & =& (\alpha \beta -\gamma (\alpha +\beta )+{\gamma }^2)(\alpha \beta +\delta (\alpha +\beta )+{\delta }^2)\\ & =& (1+\gamma +{\gamma }^2)(1-\delta +{\delta }^2)\\ & =& 1-\delta +{\delta }^2+\gamma -\gamma \delta +\gamma {\delta }^2+{\gamma }^2-{\gamma }^2\delta +{\gamma }^2{\delta }^2\\ & =& 1-\delta +\gamma +{\delta }^2-\gamma \delta +\gamma {\delta }^2+{\gamma }^2-\delta {\gamma }^2+{\gamma }^2{\delta }^2\end{array}$

Solve further as:

$\begin{array}{rcl}\left(\alpha -\gamma \right)\left(\beta +\delta \right)\left(\alpha +\delta \right)\left(\beta -\gamma \right)& =& 1-(\delta -\gamma )-\gamma \delta +\gamma \delta (\delta -\gamma )+({\gamma }^2+{\delta }^2)+1\\ & =& 1-\gamma \delta +(\delta -\gamma )(\gamma \delta -1)+7+1\\ & =& 1-1+(\delta -\gamma )(1-1)+8\\ & =& 0+0+8\\ & =& 8\end{array}$

Hence, $\left(\alpha -\gamma \right)\left(\beta +\delta \right)\left(\alpha +\delta \right)\left(\beta -\gamma \right)=8$. Thus, option (D) is correct.