Complex Numbers And Quadratic Equations question 696
Question: 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
Options:
A) p and q
B) $ \frac{1}{p} $ and $ \frac{1}{q} $
C) $ -p $ and $ -q $
D) $ p+q $ and $ p-q $
Show Answer
Answer:
Correct Answer: C
Solution:
Here m and n are the roots of equation. $ ( x+p )( x+q )-k=0 $ $ x^{2}+x(p+q)+pq-k=0 $ … (i) If m and n are the roots of equation, then $ ( x-m )( x-n )=0 $
$ \therefore x^{2}-(m+n)x+mn=0 $ … (ii) Now equation (i) should be equal to equation (ii), $ ( m + n ) = - ( p + q ) and mn = pq - k $ Now, we have to find roots of $ ( x-m ) ( x-n )+k= 0 $ $ {x^{2}}-( m+n )x+mn+k=0 $ $ {x^{2}}-( p+q )x+(pq-k)+k=0 $ $ {x^{2}}+( p+q )x+( pq-k )+k=0 $ $ {x^{2}}+( p+q )x+pq=0 $ $ {x^{2}}+px+qx+pq=0 $ $ x( x+p )+q( x+p )=0 $
$ \therefore x+q=0orx+p=0 $
$ \therefore x =- q and x = -p $
BETA
