Let a, b, c be real numbers with a ≠ 0 — JEE Mathematics
Let $a, b, c$ be real numbers with $a \ne 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$, $\beta$ is a root of $a^2x^2 - bx - c = 0$, and $0 < \alpha < \beta$, show that the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that satisfies $\alpha < \gamma < \beta$.
1 Answer
Let $f(x) = a^2x^2 + 2bx + 2c$.
We are given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$, so:
$$a^2\alpha^2 + b\alpha + c = 0 \implies b\alpha + c = -a^2\alpha^2$$
Evaluate $f(\alpha)$:
$$f(\alpha) = a^2\alpha^2 + 2(b\alpha + c) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$$
Since $a \ne 0$ and $\alpha > 0$, $f(\alpha) < 0$.
We are given that $\beta$ is a root of $a^2x^2 - bx - c = 0$, so:
$$a^2\beta^2 - b\beta - c = 0 \implies b\beta + c = a^2\beta^2$$
Evaluate $f(\beta)$:
$$f(\beta) = a^2\beta^2 + 2(b\beta + c) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$$
Since $a \ne 0$ and $\beta > 0$, $f(\beta) > 0$.
Since $f(x)$ is a continuous polynomial function and $f(\alpha) < 0$ while $f(\beta) > 0$, by the Intermediate Value Theorem, there must exist at least one root $\gamma \in (\alpha, \beta)$ such that $f(\gamma) = 0$.