Problems from Systems of inequations with one variable

Development:

We will solve both inequations independently and next we will intersect the regions of solutions: $$x(x-2)<0\Rightarrow \{\begin{array}{l}x<0\\ x-2>0\end{array}\text{ o bé }\{\begin{array}{l}x>0\\ x-2<0\end{array}$$ we must take the second option since the first one has no solution. Then : $0<x<2$.

On the other hand: $$x^2+x>-(1+x)\Rightarrow x^2+2x+1>0$$

We solve the quadratic equation $$x^2+2x+1=0\Rightarrow x={\displaystyle \frac{-2\pm \sqrt{4-4}}{2}}=-1$$

Consequently we have: $$x^2+2x+1=(x+1{)}^2=(x+1)(x+1)>0\Rightarrow x+1>0\text{ o bé }x+1>0$$

we deduce that the solutions will be: $x+1>0$ and $x+1>0$

Now we intersect the regions of both solutions and obtain the region: $$0<x<2$$

Solution:

$0<x<2$

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