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Integral Roots of x^2 - abx + a^2 + b = 0: How Many Equations? | JEE

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

Question
If α,β\alpha, \beta are the integral roots of the quadratic equation
x2−αβ x+α2+β=0,x^{2} - \alpha\beta\,x + \alpha^{2} + \beta = 0 ,
 then the number of possible different quadratic equation(s) is
A00
B11correct
C22
D44
Solution
Step 1: Apply the sum and product of roots. The roots are α\alpha and β\beta, so
α+β=αβ...(i)\alpha + \beta = \alpha\beta \qquad \text{...(i)}
αβ=α2+β...(ii)\alpha\beta = \alpha^{2} + \beta \qquad \text{...(ii)}
 Step 2: Turn (i) into a factorised form. Move everything to one side and subtract 11 from both sides:
α+β−αβ=0  ⟹  α+β−αβ−1=−1.\alpha + \beta - \alpha\beta = 0 \implies \alpha + \beta - \alpha\beta - 1 = -1 .
 The left side is −(αβ−α−β+1)=−(α−1)(β−1)-\left(\alpha\beta - \alpha - \beta + 1\right) = -(\alpha-1)(\beta-1), so
(α−1)(β−1)=1.(\alpha - 1)(\beta - 1) = 1 .
 Step 3: Solve in integers. Two integers multiply to 11 only when both are 11 or both are −1-1:
α−1=β−1=1  ⟹  α=β=2,\alpha - 1 = \beta - 1 = 1 \implies \alpha = \beta = 2 ,
α−1=β−1=−1  ⟹  α=β=0.\alpha - 1 = \beta - 1 = -1 \implies \alpha = \beta = 0 .
 Step 4: Test each pair in (ii) - this is the condition we have not used yet.
α=β=2:αβ=4,α2+β=4+2=6.4≠6\alpha = \beta = 2: \quad \alpha\beta = 4, \quad \alpha^{2}+\beta = 4 + 2 = 6 . \quad 4 \ne 6
α=β=0:αβ=0,α2+β=0+0=0.\alpha = \beta = 0: \quad \alpha\beta = 0, \quad \alpha^{2}+\beta = 0 + 0 = 0 . \quad
 Step 5: Write down the surviving equation. With α=β=0\alpha = \beta = 0 the equation becomes
x2−0⋅x+0=0,i.e.x2=0,x^{2} - 0\cdot x + 0 = 0, \quad \text{i.e.} \quad x^{2} = 0 ,
 whose roots are 0,00, 0 - integers, as required. Step 6: Exactly one equation survives, so the count is 11. Answer: (2).
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