每日一题：2020-05-26

每日一题: 2020-05-26

题目: $a$ 为何值时, 方程$x^2+2ax+2a^2-1=0$ 至少有一个正根?

参考思路

(1) 若方程两根都是整数, 则
\[
\left\{\begin{array}{lr} \Delta \geq 0 \\ x_1+x_2>0 \\ x_1x_2>0 \end{array}\right.\Rightarrow \left\{\begin{array}{lr} -1\leq a\leq 1 \\ a\lt 0 \\ a>\frac{\sqrt{2}}{2} 或 a\lt -\frac{\sqrt{2}}{2} \end{array}\right.
\]
所以可得: $-1\leq a\lt -\frac{\sqrt{2}}{2}$

(2) 若两根一正一负, 则$x_1x_2<0\Rightarrow -\frac{\sqrt{2}}{2}\lt a\lt \frac{\sqrt{2}}{2}$.

(3) 若两根一正一零, 则
\[
\left\{\begin{array}{lr} x_1+x_2>0 \\ x_1x_2=0 \end{array}\right.\Rightarrow a=-\frac{\sqrt{2}}{2}
\]

合并(1)(2)(3)得: $-1\leq a\lt \frac{\sqrt{2}}{2}$.