JEE Main 2023 · 25 Jan · Shift 2 · numerical answer

JEE Main 2023 Mathematics question on Complex Numbers and Quadratic Equations

Let a∈Ra \in \mathbb{R}, and let α,β\alpha, \beta be the roots of the equation x2+6014x+a=0x^2 + 60^{\frac{1}{4}}x + a = 0. If α4+β4=−30\alpha^4 + \beta^4 = -30, then the product of all possible values of aa is _______
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Answer: 45

Solution

Using Vieta's formulas: α + β = -60^(1/4), αβ = a.

Compute α² + β² = (α+β)² - 2αβ = √60 - 2a.

Compute α⁴ + β⁴ = (α²+β²)² - 2(αβ)² = (√60 - 2a)² - 2a² = -30.

Expanding: 60 - 4a√60 + 4a² - 2a² = -30, so 2a² - 4a√60 + 90 = 0, giving a² - 2a√60 + 45 = 0.

By Vieta's for this quadratic in a, the product of all possible values of a = 45.

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