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Q.
If $c, d$ are the roots of the equation $(x-a)(x-b)-k=0$, then the roots of the equation $(x-c)(x-d)+k=0$ are
Complex Numbers and Quadratic Equations
Solution:
We have, $(x-a)(x-b)-k=0$
$\Rightarrow x^{2}-(a+b) x+a b-k=0$
Since the roots of Eq. (1) are $c$ and $d$
$\therefore c+d=a+b, $
and $c d=a b-k$
Now $(x - c) (x - d) + k = 0$
$\Rightarrow x^{2}-(c+d) x+c d+k=0 $
$\Rightarrow x^{2}-(a+b) x+a b=0$
[Putting the values of $a+b$ and $a b$ from Eqs (2) and (3)]
$\Rightarrow (x-a)(x-b)=0 $
$\Rightarrow x=a, b .$