Quadratic Equation find the value of $\lambda$ when other roots are given in restriction

Question

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..

Answer 1

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).

Answer 2

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.

Answer 3

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 $