Question

$\alpha ,\beta$ are roots of $x^2+ax+1=0$ and $\gamma ,\delta$ are roots of $x^2+bx+1=0$. Find the value of $\left(\gamma -\alpha -\beta +\delta \right)\left(\gamma +\alpha +\beta +\delta \right)$.

• A. $a^2+b^2$
• B. $b^2-a^2$
• C. $a+b$
• D. $a-b$

Answer 1

The correct option is B

$b^2-a^2$

Step 1: Sum and product of roots of the given equations:

$\alpha ,\beta$ are roots of $x^2+ax+1=0$ . Thus:

$\begin{array}{rcl}\alpha +\beta & =& -a\\ \alpha \beta & =& 1\end{array}$

Also, $\gamma ,\delta$ are the roots of $x^2+bx+1=0$. So:

$\begin{array}{rcl}\gamma +\delta & =& -b\\ \gamma \delta & =& 1\end{array}$

Step 2: Solve the given expression:

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

Step 3: Final answer.

Therefore, the value of $\left(\gamma -\alpha -\beta +\delta \right)\left(\gamma +\alpha +\beta +\delta \right)=b^2-a^2$.

Hence, the correct option is (B).