<s>[INST] Constraints on the roots to a quadratic equation If both roots of the quadratic equation 
$$2x^2 +kx -(k+1)=0$$
are greater than $1$, then $k$ lies in what interval?
I tried to solve this using using different graphical and algebraic method but i seem to miss a crucial insight which is stopping me from getting the answer. Please help here.
 [/INST] Let $x_1,x_2$ be the two roots of your equation, so $x_1 \geq 1$, $x_2 \geq 1$.
In general, given a quadratic equation $ax^2+bx+c=0$ you have that $\frac{c}{a}$ is the product of the two roots and that $-\frac{b}{a}$ is the sum of the two roots.
In our case $x_1+x_2=-\frac{k}{2}$ and $x_1 x_2=-\frac{k+1}{2}$ and $x_1+x_2 \geq 2$ and $x_1x_2 \geq 1$.
We get $-\frac{k}{2} \geq 2$ or equivalently $k \leq -4$ and $-\frac{k+1}{2}\geq 1$ or equivalently $k \leq -3$. Together they give you $k \leq -4$.
You have to check one last thing: you equation must have two real roots, so $\Delta \geq 0$. Here $\Delta=k^2+4\cdot 2 (k+1)=k^2+8k+8$. Now $k^2+8k+8 \geq 0$ and we have that $k^2+8k+8=(k-2\sqrt{2}+4)(k+2\sqrt{2}+4)$ so $k \leq -2\sqrt{2}-4$ or $k  \geq 2\sqrt{2}-4$. 
Now we can combine our conditions and the final answer is $k \in (-\infty, -2\sqrt{2}-4]$
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