equation $7x^2-(k+13)x +k^2-k-2=0$. To find the range values of $k$

Question

The quadratic equation $7x^2-(k+13)x+k^2-k-2=0$, where $k$ is a constant, has two distinct real roots $\alpha$ and $\beta$. Given that $0<\alpha<1$ and $1<\beta<2,$ find the range of values of k.

I got $−2.055<k<4.055,$ but is unable to link to the alpha and beta.

Answer 1

Let $f(x)= 7x^2-(k+13)x+k^2-k-2.$

Now, solve the following system. $$f(0)>0,$$ $$f(1)<0$$ and $$f(2)>0,$$ which is $$k^2-k-2>0$$ $$k^2-2k-8<0$$ and $$k^2-3k>0$$ Can you end it now?

I got $$(-2,-1)\cup(3,4)$$

Answer 2

$7x^2-(k+13)x+k^2-k-2 = 0$ $\alpha, \beta \in x$

$0 < \alpha < 1$, $\alpha$ is $0→1$

$1 < \beta < 2$, $\beta$ is $1 → 2$

From this we can see than $\beta > \alpha$

Since $x$ is distinct and real, therefore it's discriminate $\delta$ is greater than $0$, $ \delta > 0$

$(k+13)^2 - 4\cdot{7}\cdot(k^2-k-2) > 0$

$k^2+2\cdot{13}\cdot{k}+13^2-28\cdot{k^2}+28\cdot{k}+28\cdot{2} > 0$

$-27k^2+54k+225 > 0$ $-3k^2+6k+25 > 0$

$3k^2-6k-25 < 0$

$\frac{3-\sqrt{84}}{3} < k < \frac{3+\sqrt{84}}{3}$

We know that discriminate, $\delta = (\alpha-\beta)^2$

Since $x$ is positive $(\beta-\alpha)^2 > 0$, $\beta - \alpha > 0$

But $\alpha + \beta = \frac{k+13}{7}$

$\beta > \frac{k+13}{7*2}$ and $\alpha < \frac{k+13}{7*2}$

I'll better write it as

$\alpha < \frac{k+13}{14} < \beta$