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How do you solve the inequality $$\sqrt{x+2}\geq{x}?$$

Now since ${x+2}$ is under the radical sign, it must be greater than or equal to ${0}$ to be defined.

So,

${x+2}\geq{0}$

Thus ${x}\geq{-2}$

Now keeping this in mind, we can solve the inequality by squaring both the sides:

${x+2}\geq{x^2}$

So ${x^2-x-2}\leq{0}$

Solving, ${(x-2)(x+1)}\leq{0}$

Therefore ${x}$ belongs to the interval ${[-1,2]}$.

As ${x}\geq{-2}$, the function is also defined.

Why does the answer say that ${x}$ belongs to ${[-2,2]}$, then?

Please feel free to point out the mistakes.

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Be careful : When you square, the inequality preserves its sign direction if both sides are positive.

Note that $\sqrt{x+2}$ is defined for $x \geq - 2$, so first you need to consider $x \geq 0$ and work as such :

$$\sqrt{x+2} \geq x \Rightarrow x+2 \geq x^2 \Leftrightarrow x^2-x-2 \leq 0 \Leftrightarrow (x-2)(x+1) \leq 0$$

This indeed yields $x \in [-1,2]$ if you also consider the negative values for which the derived inequality is satisfied .

But if $x$ is negative $(-2 \leq x < 0)$, then the (positive) square root will always be bigger than the negative left-hand side. Thus, $[-2,0)$ will do the trick in that case.

Concluding : $\sqrt{x+2} \geq x \implies x \in [-2,2]$.

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  • $\begingroup$ But we usually take the intersection of the intervals that define the radical function (here ${[-2,∞)}$) and the intervals obtained after squaring LHS and RHS (here ${[-1,2]}$) and it works pretty well in most of the examples. Is there more specific 'rule' or method for such cases? $\endgroup$ – Parth Amritkar May 18 at 12:31
  • $\begingroup$ The inequality $x^2 - x - 2 \leq 0 \implies x \in [-1, 2]$. However, since $x \geq 0$, you should have $x \in [0, 2]$. $\endgroup$ – N. F. Taussig May 19 at 9:56
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Clearly $x\geq -2$.

  • If $x<0$ then each $x\in [-2,0) $ is a solution (since negative number is always smaller than square root).
  • Now if $x\geq 0$ then you can square it, so you get $$x^2-x-2 = (x-2)(x+1)\leq 0$$ So in this case every $x\in[0,2]$ is a solution.

So finally, every $x\in [-2,2]$ is a solution.

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Once you know $x \geqslant -2$, consider first $x \in [-2, 0)$. The LHS is defined and non-negative, while the RHS is __________.

Next, consider the case $x \geqslant 0$, where you can freely square as you have done. Here you should get $x \in [0, 2]$ as the solution.

Now the solution set is the union of these cases.

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Hint:

The inequation $\;\sqrt A\ge B$, on its domain (defined by the condition $A\ge 0$) is equivalent to $$A\ge B^2\quad\textbf{ or }\quad B\le 0.$$

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  • $\begingroup$ It is a general solution. But are you sure that $B\leq0$ rather than $B<0$? I know that it does not matter actually because of union. $\endgroup$ – Artificial Stupidity May 21 at 11:27
  • $\begingroup$ @ArtificialOdorlessArmpit: You mean it's one of the basic rules on inequalities… $\endgroup$ – Bernard May 21 at 11:29
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Let $a=\sqrt{x+2}\ge0$ for real $x$

We need $$a\ge a^2-2\iff0\ge a^2-a-2=(a-2)(a+1)$$

$$\iff -1\le a\le2\ \ \ \ (1)$$

But we need to honor $a\ge0\ \ \ \ (2)$

Find the intersection of $(1),(2)$

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  • $\begingroup$ $$\implies0\le x+2\le2^2$$ $\endgroup$ – lab bhattacharjee May 18 at 13:18
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You made a mistake squaring both sides without checking the sign first. $1>-1$ but $1^2\not>(-1)^2$.

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To square both sides of an equality/inequality and obtain an equivalent statement you must be certain that both sides have the same sign (in case of inequality and both sides negative you should swich the direction of the inequality though). In your case left side is always nonnegative, therefore for nonpositive $x$ the inequality is aleays fulfiled as the $LHS \ge 0 \ge RHS$. But since the inequality doesn't make sense for $x<-2$ only nonpositive numbers greater than it satysfiy this inequality.

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