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Find a from Reciprocal Root Sum 5/12 and Product 24 | JEE Advanced

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

Question
Let α,β\alpha, \beta be the roots of x2+ax+b=0x^{2} + ax + b = 0 and γ,δ\gamma, \delta be the roots of x2−ax+b−2=0x^{2} - ax + b - 2 = 0. If
1α+1β+1γ+1δ=512andαβγδ=24,\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} = \frac{5}{12} \qquad \text{and} \qquad \alpha\beta\gamma\delta = 24 ,
 then find the value of aa.
Solution
Answer: 5
Step 1: Write down the sums and products for both equations.
α+β=−a,αβ=b,\alpha + \beta = -a, \qquad \alpha\beta = b ,
γ+δ=a,γδ=b−2.\gamma + \delta = a, \qquad \gamma\delta = b - 2 .
 Step 2: Use the product condition first.
αβγδ=b(b−2)=24....(i)\alpha\beta\gamma\delta = b(b-2) = 24 . \qquad \text{...(i)}
 Step 3: Convert the sum of reciprocals into these quantities. Pairing them off,
1α+1β=α+βαβ=−ab,1γ+1δ=γ+δγδ=ab−2.\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-a}{b}, \qquad \frac{1}{\gamma}+\frac{1}{\delta} = \frac{\gamma+\delta}{\gamma\delta} = \frac{a}{b-2} .
 Step 4: Add them and set equal to 512\dfrac{5}{12}.
a(1b−2−1b)=512.a\left(\frac{1}{b-2} - \frac{1}{b}\right) = \frac{5}{12} .
 Step 5: Combine the bracket over a common denominator.
1b−2−1b=b−(b−2)b(b−2)=2b(b−2).\frac{1}{b-2} - \frac{1}{b} = \frac{b - (b-2)}{b(b-2)} = \frac{2}{b(b-2)} .
 Step 6: Substitute (i), which says b(b−2)=24b(b-2) = 24 - the point of doing the product condition first.
a⋅224=512  ⟹  a12=512  ⟹  a=5.a \cdot \frac{2}{24} = \frac{5}{12} \implies \frac{a}{12} = \frac{5}{12} \implies a = 5 .
 Step 7: Check. From (i), b2−2b−24=0b^{2}-2b-24 = 0 gives b=6b = 6 or b=−4b = -4. With a=5, b=6a = 5,\ b = 6: −56+54=−10+1512=512\dfrac{-5}{6} + \dfrac{5}{4} = \dfrac{-10+15}{12} = \dfrac{5}{12} ✓, and 6×4=246 \times 4 = 24 . Answer: 55 (i.e. 5.005.00).
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