Quadratic Equation find the value of $\lambda$ when other roots are given in restriction Problem : 
If $\lambda$ be an integer and $\alpha, \beta$ be the roots of $4x^2-16x+\lambda$=0 such that $ 1 < \alpha <2$ and $2 < \beta <3$, then find the possible values of $\lambda$ 
My approach : 
The roots $\alpha, \beta = \frac{16 \pm \sqrt{256-16\lambda}}{8}$
$\Rightarrow \alpha, \beta  = \frac{4 \pm \sqrt{16-\lambda}}{2}$
$\Rightarrow \alpha, \beta  = \frac{4 \pm \sqrt{16-\lambda}}{2}$
$1 < \frac{4 \pm \sqrt{16-\lambda}}{2} < 2$ 
Also $ 2 < \frac{4 \pm \sqrt{16-\lambda}}{2} < 3$
Please suggest further..Thanks..
 A: The sum of the roots is $4$ and you're asking for the possible values of their product. If one root $\alpha$ varies from $1$ to $2$, we're looking at $f(\alpha)=\alpha(4-\alpha)=-(\alpha-2)^2+4$, which has its maximum at $2$. So $\lambda=4f(\alpha)$ varies from $12$ to $16$ on the given interval.
A: Hint: Since we are assuming the roots are real, we can take the minus sign for $\alpha$ and the plus sign for $\beta$.  This gives us the two inequalities:
$$2<4-\sqrt{16-\lambda}<4\\4<4+\sqrt{16-\lambda}<6$$
Now, find $\lambda$ that satisfies both of these inequalities (notice that they are actually the same inequality).
A: Using Vieta's formulas,  $\alpha+\beta=4$ and $\alpha \beta=\frac{\lambda}4\implies \lambda=4\alpha \beta $
If $1< \alpha<2 \iff 1< 4-\beta <2\iff -1>\beta-4>-2\iff 3>\beta>2$
So, one condition implies the other
Now, $\lambda=4\alpha \beta=4\alpha(4-\alpha)=16-(2\alpha-4)^2$ 
As  $1< \alpha<2 \implies 0>2\alpha-4>-2 \implies0<(2\alpha-4)^2<4 $
$\implies 0>-(2\alpha-2)^2>-4 \implies 16>16-(2\alpha-2)^2>12 $
