Answer to Question #223090 in Calculus for Akinibom Sandra

Question

Solve the inequality

x>√(x+2)

IxI / Ix+1I -1 > 1

Answer 1

1.

$$x>\sqrt{x+2}$$

We know that

$$\begin{matrix} x+2\geq0 \\ \\ \sqrt{x+2}\geq0 \end{matrix}$$

Then we have $x>0$

$$x^2>x+2$$

$$x^2-x-2>0$$

$$(x+1)(x-2)>0$$

$$x<-1\ or \ x>2$$

Since $x>0,$ then we take $x>2.$

Answer: $x>2.$

$x\in(2, \infin).$

2.

$$\dfrac{|x|}{|x+1|-1} > 1$$

$$|x+1|-1\not=0=>x\not=0, x\not=1$$

$$\begin{cases} |x+1|-1>0 \\ |x|>|x+1|-1 \end{cases}$$

$$\begin{cases} |x+1|>1 \\ |x|+1>|x+1| \end{cases}$$

$x\leq-1, x\not=-2$

$$\begin{cases} -x-1>1 \\ -x+1>-x-1 \end{cases}$$

$$\begin{cases} x<-2 \\ 2>0 \end{cases}$$

$x\in(-\infin, -2)$

$-1\leq x<0$

$$\begin{cases} x+1>1 \\ -x+1>x+1 \end{cases}$$

$$\begin{cases} x>0 \\ 2x<0 \end{cases}$$

No solution

$x>0$

$$\begin{cases} x+1>1 \\ x+1>x+1 \end{cases}$$

$$\begin{cases} x>0 \\ 1>1 \end{cases}$$

No solution

Answer: $x<-2$

$x\in(-\infin, -2)$