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If $m$ and $n$ are the roots of the equation $(x+p)(x+q)-k=0$, then the roots of the equation $(x-m)(x-n)+k=0$ are
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The correct answer is:
$-p$ and $-q$
Here $\mathrm{m}$ and $\mathrm{n}$ are the roots of equation. $(\mathrm{x}+\mathrm{p})(\mathrm{x}+\mathrm{q})-\mathrm{k}=0$
$\mathrm{x}^{2}+\mathrm{x}(\mathrm{p}+\mathrm{q})+\mathrm{p} \mathrm{q}-\mathrm{k}=0$...(i)
If $\mathrm{m}$ and $\mathrm{n}$ are the roots of equation, then $(\mathrm{x}-\mathrm{m})(\mathrm{x}-\mathrm{n})=0$
$\mathrm{x}^{2}-(\mathrm{m}+\mathrm{n}) \mathrm{x}+\mathrm{mn}=0$...(ii)
Now equation (i) should be equal to equation (ii), $(m+n)=-(p+q) a n d m n=p q-k$
Now, we have to find roots of $(x-m)(x-n)+k=0$ $x^{2}-(m+n) x+m n+k=0$
$x^{2}+(p+q) x+(p q-k)+k=0$
$\mathrm{x}^{2}+(\mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{pq}=0$
$\mathrm{x}^{2}+\mathrm{px}+\mathrm{qx}+\mathrm{pq}=0$
$x(x+p)+q(x+p)=0$
$x+q=0$ or $x+p=0$
$\mathrm{x}=-\mathrm{q}$ and $\mathrm{x}=-\mathrm{p}$
Option $(\mathrm{c})$ is correct
$\mathrm{x}^{2}+\mathrm{x}(\mathrm{p}+\mathrm{q})+\mathrm{p} \mathrm{q}-\mathrm{k}=0$...(i)
If $\mathrm{m}$ and $\mathrm{n}$ are the roots of equation, then $(\mathrm{x}-\mathrm{m})(\mathrm{x}-\mathrm{n})=0$
$\mathrm{x}^{2}-(\mathrm{m}+\mathrm{n}) \mathrm{x}+\mathrm{mn}=0$...(ii)
Now equation (i) should be equal to equation (ii), $(m+n)=-(p+q) a n d m n=p q-k$
Now, we have to find roots of $(x-m)(x-n)+k=0$ $x^{2}-(m+n) x+m n+k=0$
$x^{2}+(p+q) x+(p q-k)+k=0$
$\mathrm{x}^{2}+(\mathrm{p}+\mathrm{q}) \mathrm{x}+\mathrm{pq}=0$
$\mathrm{x}^{2}+\mathrm{px}+\mathrm{qx}+\mathrm{pq}=0$
$x(x+p)+q(x+p)=0$
$x+q=0$ or $x+p=0$
$\mathrm{x}=-\mathrm{q}$ and $\mathrm{x}=-\mathrm{p}$
Option $(\mathrm{c})$ is correct
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