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Roots of 2(x-m)(x-n) - (x-a)(x-b) = 0 | JEE Advanced Quadratic Identity

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

Question
If the roots of the quadratic equation
(x−α)(x−β)+(x−γ)(x−δ)=0(x-\alpha)(x-\beta) + (x-\gamma)(x-\delta) = 0
 are mm and nn respectively, then the roots of the equation
2(x−m)(x−n)−(x−α)(x−β)=02(x-m)(x-n) - (x-\alpha)(x-\beta) = 0
 will be
Aα+m+γ2, β+n+δ2\dfrac{\alpha+m+\gamma}{2},\ \dfrac{\beta+n+\delta}{2}
Bγ, δ\gamma,\ \deltacorrect
Cα−m, β−n\alpha - m,\ \beta - n
Dm−α+γ, n−β+δm - \alpha + \gamma,\ n - \beta + \delta
Solution
Step 1: Find the leading coefficient of the first equation. Expanding,
(x−α)(x−β)+(x−γ)(x−δ)=2x2−(α+β+γ+δ)x+(αβ+γδ),(x-\alpha)(x-\beta) + (x-\gamma)(x-\delta) = 2x^{2} - (\alpha+\beta+\gamma+\delta)x + \left(\alpha\beta + \gamma\delta\right),
 so it is a quadratic with leading coefficient 22. Step 2: Write it in factored form. A quadratic with leading coefficient 22 and roots m,nm, n is 2(x−m)(x−n)2(x-m)(x-n), and two polynomials of the same degree with the same leading coefficient and the same roots are identical. Hence
(x−α)(x−β)+(x−γ)(x−δ)=2(x−m)(x−n)....(i)(x-\alpha)(x-\beta) + (x-\gamma)(x-\delta) = 2(x-m)(x-n) . \qquad \text{...(i)}
 Step 3: Rearrange (i) to isolate the expression we are asked about.
2(x−m)(x−n)−(x−α)(x−β)=(x−γ)(x−δ).2(x-m)(x-n) - (x-\alpha)(x-\beta) = (x-\gamma)(x-\delta) .
 Step 4: Read off the roots. The right side vanishes exactly when x=γx = \gamma or x=δx = \delta. Answer: (2).
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