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..
Best Answer
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.