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.
 A: 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 $k \in (-\infty, -2\sqrt{2}-4]$
A: HINT: Roots of the given quadratic equation are:
$$x_{1,2} = \frac{-k\pm \sqrt{k^2+4\cdot2(k+1)}}{2\cdot2} = \frac{-k\pm \sqrt{k^2+8k+8}}{4}$$
and we are asked to find interval for $k$ when both of these roots are greater than $1$. After Dr. Sonnhard Graubner's comment, I realized that you should also check for $k^2+8k+8 \ge 0$. This will give you one more interval for $k$. Can you take it from here?
A: Let $f(x)=2x^2+kx-(k+1)$. The parabola $y=f(x)$ is concave up (because the coefficient of $x^2$ is positive). For the roots to be greater than $1$, we need  three things: 

real roots,  $f(1)>0$ and the $x-$coordinate of the
  vertex of the parabola to be greater than $1$.

1). $k^2+8(k+1) \geq 0 \implies \color{blue}{k \geq -4+2\sqrt{2} \text{ or } k \leq -4-2\sqrt{2}}$.
2). $f(1)=1>0$. So nothing to be done here.
3). The vertex occurs at $x=-\frac{k}{4}$. So we want $\frac{-k}{4}>1 \implies \color{blue}{k <-4}$.
The common values of $k$ are 
$$\color{red}{k \leq -4-2\sqrt{2}}$$
A: Denote $p(x)=2x^2+kx-(k+1)$.


*

*First $p(x)$ must have real roots, i.e. its discriminant $\;\Delta(k)=k^2+8(k+1)\ge 0$. Its roots are
$$\kappa_1=-4-2\sqrt 2,\quad \kappa_2=-4+2\sqrt. 2$$
So the first condition is $\;k\in(-\infty, \kappa_1]\cup[\kappa_2,+\infty)$.

*Observe that $\;\Delta(1)>0$, so $1$ is outside  the the interval of the roots ($\xi_1,\xi_2$), i.e. we have either $1<\xi_1\le\xi_2$ or $\xi_1\le \xi_2<1$.

*We want to have the first case. This is equivalent to
$$1<\frac{\xi_1+\xi_2}2=-\frac k4\iff k<-4.$$
Grouping these conditions, we obtain
$$k\le -4-2\sqrt 2.$$
A: If $k> 0$ then for $x> 1$
$$\therefore\,2\,x^{\,2}+ kx- k- 1> 2+ k- k- 1> 0$$
If $-\,4\leqq k\leqq 0$ then for $x> 1$
$$\therefore\,2\,x^{\,2}+ kx- k- 1\geqq 2\,x^{\,2}- 4(\,x- 1\,)- 1> 0$$
But $2\,x^{\,2}+ kx- k- 1= 0\,\therefore\,k< -\,4$.
Using $\lceil$ https://en.wikipedia.org/wiki/Quadratic_function $\rfloor$ we receive
$$2\,x^{\,2}+ kx- k- 1= 0$$
has two roots then
$$\therefore\,{\rm discriminant}= k^{\,2}+ 8\,k+ 8> 0$$
$$\therefore\,(\,k+ 4\,)^{\,2}> 8\,\therefore\,k< -\,4- \sqrt{8}$$
A: $$ (x_1-1)(x_2-1)>0 \implies x_1x_2-(x_1+x_2)+1>0$$
so by Vieta formulas we have $$-{k+1\over 2}+{k\over 2} +1>0\implies k\in \mathbb{R}$$
So you have to check only when the disciriminat is positive and you are done.
