Quadratic Equations
Quadratic Equations
nta_abhyas_2025
Grade None
Question:
If $\alpha$, $\beta$ and $\gamma$ are the roots of the equation $x^3 + x + 2 = 0$, then the equation whose roots are $(\alpha - \beta)(\alpha - \gamma)$, $(\beta - \alpha)(\beta - \gamma)$ and $(\gamma - \alpha)(\gamma - \beta)$ is
(\lambda)^2 - 6\lambda^2 + 216 = 0
(\lambda)^3 - 3\lambda^2 + 112 = 0
(\lambda)^2 + 6\lambda^2 - 216 = 0
None
Step-by-Step Solution
Key Concept: Use Vieta's formulas and algebraic identities to transform equations through clever variable substitutions involving roots
Given $x^3 + 3x^2 - 112 = 0$ with substitution $x = (\alpha - \beta)(\alpha - \gamma)$, we expand and use $y = \alpha^2 - \alpha(\beta + \gamma) + \frac{\beta\gamma}{2}$. From $\alpha^3 - \alpha - 2 = 0$, we derive $\alpha^2 = \alpha + 2$. Through substitution and simplification, $y = \alpha^2 - (\alpha + 1) + \frac{-(\alpha+2)}{2} = -\frac{\alpha}{2}$. Therefore $(\frac{-\alpha}{2})^2 + 3(\frac{-\alpha}{2}) - 2 = 0$, which yields $x^2 + 3x^2 - 112 = 0$.
Correct Answer: x^2 + 3x^2 - 112 = 0