Problems from Quadratic inequations

Development:

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

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

where $x_1<x_2$.

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

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 $1$. $$x^2-x<0\Rightarrow x(x-1)<0\Rightarrow$$ $$\Rightarrow \{\begin{array}{l}x>0\text{ and }x-1<0\Rightarrow x<1\\ x<0\text{ and }x-1>0\Rightarrow x>1\end{array}$$

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

Solution:

a) There is no solution.

b) The inequation is satisfied for those $x$ such that: ${\displaystyle \frac{3-\sqrt{5}}{2}}<x<{\displaystyle \frac{3+\sqrt{5}}{2}}$.

c) The inequation is satisfied for those $x$ such that: $0<x<1$.

Hide solution and development