+4 marks−1 if incorrectNumericalPrevious-year question
Relating Polynomial Roots to Power Sums via Vieta's Formulas
Let a ∈ ℝ and let α, β be the roots of the equation x^2 + 60^(1/4)x + a = 0 If α^4 + β^4 = -30, then the product of all possible values of a is
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Hint 1 of 4
Using Vieta's formulas for x^2 + 60^(1/4)x + a = 0, what are the expressions for the sum and product of the roots?
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Correct answer
The product of all possible real values of a is -45.
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Solution
StepWorking
01Given
The quadratic equation is x2+601/4x+a=0 with roots α,β and a∈R. We are given α4+β4=−30.
02Goal
Find the product of all possible real values of a.
03Approach
Apply Vieta's relations to express α+β and αβ in terms of a. Then express α2+β2 and (α2+β2)2 to relate α4+β4 to a, obtaining a quadratic equation in a. Finally, use Vieta's product of roots for a (after checking that the discriminant is non-negative).
04Execute
By Vieta's formulas:
α+β=−601/4,αβ=a
Square both sides of the sum:
α2+β2=(α+β)2−2αβ=60−2a
05Execute
Square α2+β2 to reach α4+β4:
(α2+β2)2=α4+β4+2α2β2
Substitute the known values:
(60−2a)2=−30+2a260−4a60+4a2=−30+2a22a2−460a+90=0
Divide through by 2:
a2−260a+45=0
✓Verify
Check the discriminant of a2−260a+45=0:
Da=(−260)2−4(1)(45)=240−180=60>0
Since Da>0, both values of a are real. By Vieta's formulas, their product is the constant term: Product=45. Note that α4+β4=−30<0 implies the roots α,β are complex conjugate pairs, which is entirely consistent with a∈R.
Hints that build this answer step by step
Using Vieta's formulas for x^2 + 60^(1/4)x + a = 0, what are the expressions for the sum and product of the roots?
α + β = -60^(1/4) and αβ = a
How can α^4 + β^4 be expressed in terms of (α + β) and αβ?
((α + β)^2 - 2αβ)^2 - 2(αβ)^2
Substitute α + β = -60^(1/4) and αβ = a into α^4 + β^4 = -30. Which quadratic equation in a results?
a^2 - 2√15 a - 45 = 0
From the quadratic equation 2a^2 - 8√15 a + 90 = 0 (or a^2 - 4√15 a + 45 = 0), what is the product of all possible real values of a?
Can the sum of fourth powers α4+β4 be negative for real numbers?
No, for real numbers it cannot. But the question states a∈R, not that the roots α,β are real. For complex roots, even powers can sum to a negative real number.
Do we need to solve for the individual values of a?
No, once you check that the discriminant is positive (so real values of a exist), the product of roots of a2−260a+45=0 is directly the constant term, 45.
