Theory of EquationseasyFree

Theory of Equations: Roots Value Equal

JEE Maths question with a full step-by-step solution.

Question
If α,β\alpha, \beta are the roots of x2−p(x+1)−c=0x^2 - p(x+1) - c = 0, c≠1c \neq 1, then the value of (α+1)(β+1)(\alpha+1)(\beta+1) is equal to
A1−c1-ccorrect
Bp+cp+c
C1+c1+c
D1−2p1-2p
Solution
Step 1: Rewrite the equation as x2−px−(p+c)=0x^2 - px - (p+c) = 0. By Vieta's formulas: α+β=p\alpha + \beta = p and αβ=−(p+c)\alpha\beta = -(p+c). Step 2: Expand:
(α+1)(β+1)=αβ+(α+β)+1=−(p+c)+p+1=1−c(\alpha+1)(\beta+1) = \alpha\beta + (\alpha+\beta) + 1 = -(p+c) + p + 1 = 1 - c
 Correct answer: (1)
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