Question

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

• A. c, d
• B. a, c
• C. b, d
• D. a, b

Answer 1

Answer: (D)

We have, $(x-a)(x-b)-k=0$
$\Rightarrow x^2-(a+b)x+ab-k=0$
Since the roots of Eq. (1) are $c$ and $d$
$\therefore c+d=a+b,$
and $cd=ab-k$
Now $(x-c)(x-d)+k=0$
$\Rightarrow x^2-(c+d)x+cd+k=0$
$\Rightarrow x^2-(a+b)x+ab=0$
[Putting the values of $a+b$ and $ab$ from Eqs (2) and (3)]
$\Rightarrow (x-a)(x-b)=0$
$\Rightarrow x=a,b.$