Quadratic inequations

a) $x^2+x+1 < -x \Rightarrow x^2+2x+1 < 0 $

We find the solutions of the equation $x^2+2x+1=0$: $$ x=\dfrac{-2\pm\sqrt{4-4}}{2}=-1$$ There is only one solution. So, $$x^2+2x+1 < 0 \Rightarrow (x-1)^2<0 $$ There is no solution, because the square of a number is always positive.

b) $$x(1-x)+2x-2 > -1 \Rightarrow x-x^2 +2x-2+1 > 0 \Rightarrow $$

$$\Rightarrow -x^2+3x-1 > 0 \Rightarrow x^2-3x+1 < 0$$

We find the solutions of the equation $x^2-3x+1=0$: $$ x=\dfrac{ 3\pm\sqrt{9-4}}{2}=\dfrac{3\pm\sqrt{5}}{2} \Rightarrow x_1= \dfrac{3-\sqrt{5}}{2}, \ x_2=\dfrac{3+\sqrt{5}}{2}$$

where $x_1 < x_2$.

$$x^2-3x+1 < 0 \Rightarrow (x-x_1)(x-x_2) < 0 \Rightarrow$$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ (x-x_1) > 0 \ \text{ and } \ (x-x_2) < 0 \\ \text{b) } \ (x-x_1) < 0 \ \text{ and } \ (x-x_2) > 0 \end{array} \right. $$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ x > x_1 \ \text{ and } \ x < x_2 \\ \text{b) } \ x < x_1 \ \text{ and } \ x > x_2 \end{array} \right. $$

and as $x_1 < x_2$, we have $x_1 < x < x_2 $.

c) $$(x-2)(x+1)+x > (2x-1)x -2 \ \Rightarrow \ x^2+x-2x-2+x > 2x^2-x -2 $$ $$\Rightarrow \ 0 > 2x^2-x-2-x^2 -x +2x +2 -x \ \Rightarrow \ 0 > x^2 -x $$

We find the solutions of the equation $x^2 -x=0$: $$ 0=x^2-x=x(x-1) \Rightarrow x_1=0 \ \text{ and } \ (x-1)=0 \Rightarrow x_2=1 $$

We have two solutions: $0$ and $1$. $$x^2 -x < 0 \Rightarrow x(x-1) < 0 \Rightarrow$$ $$\Rightarrow \left\{ \begin{array}{l} x > 0 \ \text{ and } \ x-1 < 0 \Rightarrow x < 1\\ x < 0 \ \text{ and } \ x-1 > 0 \Rightarrow x > 1 \end{array} \right. $$

by more restrictive inequations we obtain $x > 0$ and $x<1$.