If α,β are the roots of the equation (x−a)(x−b)+c=0(c≠0), then the roots of the equation (x−c−α)(x−c−β)=c are

Given Equation (x−a)(x−b)+c=0⇒ x2−(a+b)x+ab+c=0α,β are roots then α+β=a+b, αβ=ab+cand (x−c−α)(x−c−β)=c⇒ x2−xc−xβ−xc+c2+cβ−αx+αc+αβ=c⇒ x2−2xc−(α+β)x+c(β+α)+αβ+c2=c⇒ (x−c)2−(a+b)(x−c)+ab=0Let x−c=y⇒ y2−(a+b)y+ab=0⇒ (y−a)(y−b)=0⇒ (x−c−a)(x−c−b)=0∴ x= a+c, b+c

If α,β are the roots of the equation (x−a)(x−b)+c=0(c≠0), then the roots of the equation (x−c−α)(x−c−β)=c are

Given Equation (x−a)(x−b)+c=0⇒ x2−(a+b)x+ab+c=0α,β are roots then α+β=a+b, αβ=ab+cand (x−c−α)(x−c−β)=c⇒ x2−xc−xβ−xc+c2+cβ−αx+αc+αβ=c⇒ x2−2xc−(α+β)x+c(β+α)+αβ+c2=c⇒ (x−c)2−(a+b)(x−c)+ab=0Let x−c=y⇒ y2−(a+b)y+ab=0⇒ (y−a)(y−b)=0⇒ (x−c−a)(x−c−b)=0∴ x= a+c, b+c

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$I f α, β a r e t h e r o o t s o f t h e e q u a t i o n (x - a) (x - b) + c = 0 (c \neq 0), t h e n t h e r o o t s o f t h e e q u a t i o n (x - c - α) (x - c - β) = c a r e$

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$a + c a n d b$

$a - c a n d b - c$

$a + c a n d b + c$

$a a n d b + c$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n (x - a) (x - b) + c = 0 \\ \Rightarrow x^{2} - (a + b) x + a b + c = 0 \\ α, β a r e r o o t s t h e n α + β = a + b, α β = a b + c \\ a n d (x - c - α) (x - c - β) = c \\ \Rightarrow x^{2} - x c - x β - x c + c^{2} + c β - α x + α c + α β = c \\ \Rightarrow x^{2} - 2 x c - (α + β) x + c (β + α) + α β + c^{2} = c \\ \Rightarrow {(x - c)}^{2} - (a + b) (x - c) + a b = 0 \\ L e t x - c = y \\ \Rightarrow y^{2} - (a + b) y + a b = 0 \\ \Rightarrow (y - a) (y - b) = 0 \\ \Rightarrow (x - c - a) (x - c - b) = 0 \\ ∴ x = a + c, b + c \end{array}$