JEE Advance - Mathematics (1989 - No. 13)
Let a, b, c be real numbers, $$a \ne 0$$. If $$\alpha \,$$ is a root of $${a^2}{x^2} + bx + c = 0$$. $$\beta \,$$ is the root of $${a^2}{x^2} - bx - c = 0$$ and $$0 < \alpha \, < \,\beta $$, then the equation $${a^2}{x^2} + 2bx + 2c = 0$$ has a root $$\gamma $$ that always satisfies
$$\gamma = {{\alpha + \beta } \over 2}$$
$$\gamma = \alpha + {\beta \over 2}$$
$$\gamma = \alpha $$
$$\alpha < \gamma < \beta $$
Comments (0)
