Quadratic Equations
Vieta's formulas and symmetric functions of roots
nta_pyq_2023_jan
Grade 11
Question:
Let $a \in \mathbb{R}$ and let $\alpha, \beta$ be the roots of the equation $x^2 + 60^{\frac{1}{4}}x + a = 0$. If $\alpha^4 + \beta^4 = -30$, then the product of all possible values of $a$ is ______.
Step-by-Step Solution
Key Concept: Use Vieta's formulas: $\alpha + \beta = -60^{1/4}$, $\alpha\beta = a$. Express $\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2\alpha^2\beta^2 = [(\alpha+\beta)^2 - 2\alpha\beta]^2 - 2a^2$.
$\alpha^4+\beta^4 = (60^{1/2}-2a)^2 - 2a^2 = -30$. Expanding: $60 + 4a^2 - 4a\cdot60^{1/2} - 2a^2 = -30 \Rightarrow 2a^2 - 4\cdot60^{1/2}a + 90 = 0$. Product of all possible values of $a = \frac{90}{2} = 45$.
Correct Answer: 45