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)
$\therefore x-c=a$
$x-c=b$
$x=a+c$
$x=b+c$