Let α, β, γ be the roots of (x-a)(x-b)(x-c)=d, d ≠ 0, then the roots of the equation (x-α)(x-β)(x-γ)+d=0 are :

Question

Let α, β, &gamma; be the roots of (x-a)(x-b)(x-c)=d, d ≠ 0, then the roots of the equation (x-α)(x-β)(x-&gamma;)+d=0 are : (A) a+1, b+1, c+1 (B) a, b, c (C) a-1, b-1, c-1 (D) ( a / b ), ( b / c ), ( c / a )

Tardigrade Admin · Accepted Answer

Clearly (x-a)(x-b)(x-c)-d =(x-α)(x-β)(x-&gamma;) ∴ if α, β, &gamma; are the roots of given equation then (x-α)(x-β)(x-&gamma;)+d=0 will have roots a, b, c.

Q. Let $α, β, γ$ be the roots of $(x - a) (x - b) (x - c) = d, d \neq = 0$ , then the roots of the equation $(x - α) (x - β) (x - γ) + d = 0$ are :

$a + 1, b + 1, c + 1$

$a, b, c$

$a - 1, b - 1, c - 1$

$\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$

Clearly $(x - a) (x - b) (x - c) - d$
$= (x - α) (x - β) (x - γ)$
$∴$ if $α, β, γ$ are the roots of given equation
then $(x - α) (x - β) (x - γ) + d = 0$
will have roots $a, b, c .$

Q. Let α,β,γ be the roots of (x−a)(x−b)(x−c)=d,d=0, then the roots of the equation (x−α)(x−β)(x−γ)+d=0 are :

a+1,b+1,c+1

a,b,c

a−1,b−1,c−1

ba​,cb​,ac​