<s>[INST] 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..
 [/INST] 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 $
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