Solving inequalities involving square roots
What is the correct way to solve inequalities like $\sqrt {x+2}>x $ ?
I deduced that $x$ must be greater than$-2$ and also on squaring I get $-1 \leq x \leq 2$ but even on combining these two I'm not getting the correct answer.What am I missing?
$\endgroup$ 34 Answers
$\begingroup$First note that $x$ must be greater than $-2$, because of the square root. Then you have two cases :
If $x \in [-2,0[$ the inequality is true since a square root is positive.
If $x \in [0,+\infty)$, then both sides are positive and you can square it to obtain the equivalent inequation $$x+2 > x^2.$$ That is $$x^2-x-2 < 0.$$ The discriminant here is $9$ and you can hence deduce the factorisation $$(x-2)(x+1) <0.$$ Then do a sign table with the two factors to conclude.
$\endgroup$ $\begingroup$Observe if $-2 \leq x \leq 0$, the inequality holds trivially. If $x > 0$, square both sides to get an equivalent one: $x+2 > x^2 \iff x^2 -x - 2 < 0\iff (x-2)(x+1) < 0\iff -1 <x < 2\implies -2 \leq x < 2$
$\endgroup$ $\begingroup$Beware that
$$a>b$$ does not imply $$a^2>b^2.$$
Indeed,
$$a^2-b^2=(a-b)(a+b)>0$$ is equivalent to $a-b>0$ only when $a+b>0$, otherwise the inequality gets reversed.
$\endgroup$ $\begingroup$Use: $$\sqrt{f(x)}>g(x) \Leftrightarrow$$
$\Leftrightarrow \begin{cases} g(x)<0\\ f(x)\ge0\end{cases}$ or $\begin{cases} g(x)\ge0\\ f(x)>\left(g(x)\right)^2 \end{cases}$
$\endgroup$!$$\sqrt{x+2}>x$$
!Case 1) $$\begin{cases} x<0\\ x+2\ge0\end{cases}$$
!Case 2) $$\begin{cases} x\ge0\\ x+2>\left(x \right)^2 \end{cases}$$
