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.
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 $$4(2a^2+6a+1)>0.$$
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: $\;a<-3\;\text{ or }\; a>0$.
