What are the values ​of $a$, one of the roots of the equation is greater than $1$ and the other is less than $1$?

Question

What are the values ​​of $a$, one of the roots of the equation is greater than $1$ and the other is less than $1$? $$(2a+1)x^2-ax+a-2=0$$

I tried Vieta's formulas for quadratic equation, but I could not get the correct solution:

$x_1+x_2= \frac {a}{2a+1}, x_1×x_2= \frac {a-2}{2a+1}$

Answer 1

Rewrite the question by shifting the x-axis by 1 to the right (substitute $x=y-1$) so that one root is less that 0 and the other one is greater than 0, and Proceed in your way. The second Vieta's formula will look better.

Answer 2

Distinguish cases $2a+1\lt 0$, $=0$ (impossible given there are two roots), $\gt 0$.

• If $2a+1\gt 0$, the statement is equivalent to the statement that the quadratic function $(2a+1)x^2-ax+a-2$ is negative at $x=1$, i.e. $2a-1\lt 0$, so we have $a\in (-1/2, 1/2) $.

• If $2a+1\lt 0$, the statement is equivalent to the statement that the quadratic function $(2a+1)x^2-ax+a-2$ is positive at $x=1$, i.e. $2a-1\gt 0$. This has no solutions.

The overall solution is: $a\in (-1/2, 1/2) $.