Find all values of $a$ for which the equation $2x^2 - 2(2a + 1)x + a(a-1)=0$ has roots $\alpha$ and $\beta$ satisfying the condition $\alpha \lt a \lt \beta$.
My attempt : I have observed the coefficient of $x^2$ here is $2 \gt 0$. I tried to evaluate the determinant which I obtained as $4(2a^2+6a+1)$, which is not helping me to proceed further. Any help will be appreciated.
On
This means $P_a(x)=2x^2-2(2a+1)x+a(a-1)$ has two real roots, so the reduced discriminant $$\Delta'(a)=(2a+1)^2-2a(a-1)=2a^2+6a+1 >0,$$ and $a\,$ separates the roots of $P_a(x)$, i.e. $$2P_a(a)<0\iff -a^2-3a<0\iff a(a+3)>0\iff a\in (-\infty,-3)\cup(0,+\infty).$$ Now both roots of $\Delta'(a)$ are negative and $>-3$ since the smallest root is $$\dfrac{-3-\sqrt 7}2>\dfrac{-3-3}2=-3.$$ Combining these results, we obtain as the solution: $\;a<-3\;\text{ or }\; a>0$.
The thing to notice is that the quadratic formula tells you what you're two (possibly repeated) roots are going to be. For real roots, which you need to have $\alpha<a<\beta$, the discriminant must be positive. Moreover, the least root \alpha will be of the form $c-d\sqrt\Delta$, where $\Delta$ is your discriminant, and the greater root $\beta$ will be of the form $c+d\sqrt\Delta$. You then have a system of two inequalities, $\alpha<a$ and $\beta>a$.
In particular, you need to solve the following two inequalities: $$a+\frac12-\frac14\sqrt{4(2a^2+6a+1)}<a$$ $$a+\frac12+\frac14\sqrt{4(2a^2+6a+1)}>a$$
There is one more implicit inequality for the solutions to make sense. Since your answers must be distinct reals, you need your discriminant positive $$4(2a^2+6a+1)>0.$$