Question

If $\alpha, \beta$ are roots of the equation $(x-a) (x-b) +c=0 (c \ne 0)$, then roots of the equation $(x-c-\alpha) (x-c-\beta)=c$ are

• A. $a$ and $b + c$
• B. $a^{2}+c^{2}$ and $b^{2}+c^{2}$
• C. $a+c$ and $a-c$
• D. $a+c$ and $b+c$

Answer 1

Answer: (D)

Given, $(x-a) (x-b) +c=0$, have root $\alpha, \beta$
$\therefore x^{2}-x (a+b) + (ab+c)=0$
$\therefore S = \alpha +\beta = a+b , P = \alpha \beta = ab +c$
Now to find the roots of the equation
$(x-c-\alpha) (x-c-\beta) = c$
i.e., $(x^{2}+c^{2}-2xc)-(x-c) (a+b) +ab+c=c$
i.e., $(x-c)^{2}-(x-c) (a+b) +ab +c=c$
(quadratic in x - c)
i.e., $(x-c-a) (x-c-b)=0$
$\therefore x=a+c, b+c$ are required roots
Short Cut Method :
Fact $x^{2}-(a+b) x+ab =0$
having roots a, b i.e. value of x are a and b.
$\therefore (x-c)^{2}-(x-c) (a+b) +ab =0$
(quadratic in x - c)
$\therefore x-c=a$
$x-c=b$
$x=a+c$
$x=b+c$