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Let $\mathrm{a}, \mathrm{b}$ be the solutions of $\mathrm{x}^2+\mathrm{px}+1=0$ and $\mathrm{c}$, $\mathrm{d}$ be the solution of $\mathrm{x}^2+\mathrm{qx}+1=0$. If $(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})$ and $(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})$ are the solution of $\mathrm{x}^2+\mathrm{ax}+\beta=0$, then $\beta$ is equal to
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The correct answer is:
$q^2-p^2$
Since, $a+b=-p, a b=1$ $\qquad$ ...(i)
and $c+d=-q, c d=1$
Now, $(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})$ and $(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})$ are the roots of $\mathrm{x}^2+\mathrm{ax}+\beta=0$
$(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})=\beta$
$\Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta$
$\Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta$
$\Rightarrow\left(1+\mathrm{pc}+\mathrm{c}^2\right)\left(1-\mathrm{pd}+\mathrm{d}^2\right)=\beta$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 2+d^2+c^2-p^2=\beta$
$\Rightarrow 2 c d+c^2+d^2-p^2=\beta$
$(\mathrm{c}+\mathrm{d})^2-\mathrm{p}^2=\beta$
$\mathrm{q}^2-\mathrm{p}^2=\beta[\because(\mathrm{c}+\mathrm{d})=-\mathrm{q}]$
and $c+d=-q, c d=1$
Now, $(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})$ and $(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})$ are the roots of $\mathrm{x}^2+\mathrm{ax}+\beta=0$
$(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})=\beta$
$\Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta$
$\Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta$
$\Rightarrow\left(1+\mathrm{pc}+\mathrm{c}^2\right)\left(1-\mathrm{pd}+\mathrm{d}^2\right)=\beta$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 2+d^2+c^2-p^2=\beta$
$\Rightarrow 2 c d+c^2+d^2-p^2=\beta$
$(\mathrm{c}+\mathrm{d})^2-\mathrm{p}^2=\beta$
$\mathrm{q}^2-\mathrm{p}^2=\beta[\because(\mathrm{c}+\mathrm{d})=-\mathrm{q}]$
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