A) p and q
B) \[\frac{1}{p}\] and \[\frac{1}{q}\]
C) \[-p\]and \[-q\]
D) \[p+q\] and \[p-q\]
Correct Answer: C
Solution :
Here m and n are the roots of equation. \[\left( x+p \right)\left( x+q \right)-k=0\] \[{{x}^{2}}+x(p+q)+pq-k=0\] ... (i) If m and n are the roots of equation, then \[\left( x\,-m \right)\left( x-n \right)=0\] \[\therefore \,\,\,\,\,{{x}^{2}}-(m+n)x+mn=0\] ... (ii) Now equation (i) should be equal to equation (ii), \[\left( m + n \right) = - \left( p + q \right) and mn = pq - k\] Now, we have to find roots of \[\left( x\,-m \right) \left( x-n \right)+k= 0\] \[{{\operatorname{x}}^{2}}-\left( m+n \right)x+mn+k=0\] \[{{\operatorname{x}}^{2}}-\left( p+q \right)x+(pq-k)+k=0\] \[{{\operatorname{x}}^{2}}+\left( p+q \right)x+\left( pq\,-k \right)+k=0\] \[{{\operatorname{x}}^{2}}+\left( p+q \right)x+pq=0\] \[{{\operatorname{x}}^{2}}+px+qx+pq=0\] \[\operatorname{x}\left( x+p \right)+q\left( x+p \right)=0\] \[\therefore \,\,\,\,\,\operatorname{x}+q=0\,\,or\,\,x+p=0\] \[\therefore \,\,\,\,\,\,\operatorname{x} =- q \,and \,x = -p\]
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