Let α ∈ℝ and let α,β be the roots of the equation x² + 60^(1/4)x + a = 0. If α ⁴ + β ⁴ = - 30, then the product of all possible values of a is ____________.

Solution

Maths sir · 4 steps

Build α⁴+β⁴ by squaring twice: go up through α²+β² and you get a quadratic in a.

1. Sum and product. From the equation, α+β = -60^(1/4) and αβ = a. So (α+β)² = 60^(1/2) = √60 and α²+β² = √60 - 2a.

   α⁴+β⁴ = (α²+β²)² - 2α²β² = (√60-2a)² - 2a²

2. Expand and equate. This gives 60 - 4√60 a + 4a² - 2a² = 2a² - 4√60 a + 60. Setting it equal to -30 gives 2a² - 4√60 a + 90 = 0, that is, a² - 2√60 a + 45 = 0.
3. Check real values. The discriminant is 4 × 60 - 4 × 45 = 60 > 0, so there are two real values of a.
4. So. The product of the two values of a is the constant term divided by the leading coefficient, 45/1 = 45.

The common slip: 22.5. Subtracting 2αβ instead of 2α²β² in the second squaring gives 4a²-(4√60+2)a+90=0 and a product of 22.5.

The idea to keep. For higher power sums, climb one step at a time: α²+β² from s²-2p, then α⁴+β⁴ from (α²+β²)²-2p². Then use Vieta again on the new equation.

Answer from the NTA final answer key. Solution by PrepWiser.