Question

Let $a$,$b$ be the solutions of $x^2+px+1=0$and $c$,$d$ be the solutions of $x^2+qx+1=0$. If $(a-c)(b-c)$ and $(a+d)(b+d)$ are the solutions of $x^2+\alpha x+\beta =0$, then $\beta$ equals

• A. $p+q$
• B. $p-q$
• C. $p^2+q^2$
• D. $q^2-p^2$

Answer 1

The correct option is D

$q^2-p^2$

Explanation for correct option:

Finding the value of $\beta$:

If $\alpha$and $\beta$be the root of the equation $a{x}^2+bx+c=0$, then $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$.

If $a$,$b$ be the solutions of the equation $x^2+px+1=0$, then $a+b=-p$ and $ab=1$

If $c$,$d$ be the solutions of the equation $x^2+qx+1=0$. then $c+d=-q$ and $cd=1$

Given,

If $(a-c)(b-c)$ and $(a+d)(b+d)$ are the solutions of $x^2+\alpha x+\beta =0$,then

$\left[(a-c)(b-c)\right]+\left[(a+d)(b+d)\right]=\alpha$ and $\left[(a-c)(b-c)\right]\times \left[(a+d)(b+d)\right]=\beta$

Finding $\beta$

$$\begin{gathered} \left[(a-c)(b-c)\right]\times \left[(a+d)(b+d)\right]=\beta \\ \left(ab-ac-bc+c^2\right)\left(ab+ad+bd+d^2\right)=\beta \\ \left(1-c\left(a+b\right)+c^2\right)\left(1+d\left(a+b\right)+d^2\right)=\beta ...\left[\because ab=1\right] \\ \left(1+pc+c^2\right)\left(1-pd+d^2\right)=\beta ...\left[\because a+b=-p\right] \end{gathered}$$

$$\begin{gathered} 1-pd+d^2+pc-p^{2}cd+pc{d}^2+c^2-c^{2}pd+c^2{d}^2=\beta \\ 1-p\left(d-c\right)+d^2-p^2\left(cd\right)+p\left(cd\right)\left(d-c\right)+c^2+1^2=\beta \\ 1-p\left(d-c\right)+p\left(1\right)\left(d-c\right)+d^2-p^2\left(1\right)+c^2+1^2=\beta ...\left[\because c+d=-q,cd=1\right] \\ 2+pq+d^2-p^2-pq+c^2=\beta \end{gathered}$$

$$\begin{gathered} 2+d^2-p^2+c^2=\beta \\ 2cd+d^2-p^2+c^2=\beta ...\left[\because cd=1\right] \\ {\left(c+d\right)}^2-p^2=\beta ...\left[\because {\left(a+b\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{b}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\right] \\ {\left(-q\right)}^2-p^2=\beta \\ {q}^2-p^2=\beta ...\left[\because c+d=-q\right] \end{gathered}$$

Hence, option (D) is the correct answer