An inequality involving ##x## on both sides: ##\sqrt{x+2}\ge x##

[brotherbobby]

Homework Statement

Solve : $\sqrt{x+2} \ge x$

Relevant Equations

For $\sqrt{f(x)}$, we must have both (1) $\sqrt{f(x)} \ge 0$ and (2) $f(x)\ge 0$

Problem statement : Let me copy and paste the problem as it appears in the text on the right.Attempt (myself) : By looking at $\large{\sqrt{x+2}\ge x}$, from my Relevant Equations above, we have the following :

1. Outcome $\mathbf{x \ge 0}$, since square roots are always positive.
2. Function inside the square root, $x+2 \ge 0\Rightarrow \mathbf{x \ge -2}$.

Armed with these inequalities, I square both sides of it. $x+2 \ge x^2\Rightarrow x^2 - x - 2 \le 0\Rightarrow (x-2)(x+1)\le 0\Rightarrow \mathbf{-1\le x \le 2}$.

Looking at the three (bold faced) solutions above and merging them, I find that the answer must be : $\boxed{\mathbf{0 \le x \le 2\Rightarrow x \in [0,2]}}$.But the book has a different answer. Attempt (Text Solution) : I copy and paste below the solution to the problem as it appears in the text.

Doubt : Of course is the text correct. For instance how can $x$ lie in the interval $[-2,2]$? That would imply that $x$ can be equal to -1, which would make the square root of a quantity negative.

A hint or suggestion would be welcome as to where have I gone wrong.

 

[pasmith]

Your error is here:

1. Outcome $\mathbf{x \ge 0}$, since square roots are always positive.

It is indeed the case that $\sqrt{x + 2} \geq 0$ for $x \geq - 2$, but this admits the possibility that $$\sqrt{x + 2} \geq 0 \geq x.$$ Thus $x \in [-2,0]$ satisfies the inequality.

 

[Gavran]

Different approach:

$\sqrt { x + 2 } = t$
$x = t ^ 2 - 2$
$t \geq t ^ 2 - 2$
$t ^ 2 - t - 2 \leq 0$
$t \in [-1, 2]$ and because $t \geq 0$
$t \in [0, 2]$
$\sqrt { x + 2 } \geq 0$ and $\sqrt { x + 2 } \leq 2$
$x + 2 \geq 0$ and $x + 2 \leq 4$
$x \geq -2$ and $x \leq 2$
$x \in [-2, 2]$

 

[Mark44]

Thus $x \in [-2,0]$ satisfies the inequality.

As does the interval [0, 2], which I believe you knew but didn't state.

If you graph $y = \sqrt{x + 2}$ and $y = x$ on the same coordinate axis system, you can see that $y = \sqrt{x + 2}$ lies above the graph of y = x for $x \in [-2, 2)$ and meets it at x = 2.