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If the inequality $ (x+2)^{\frac{1}{2}} > x $ is satisfied. what is the range of x ?

My approach - I squared both the sides and proceeded on to solve the quadratic obtained in order to solve the inequality. However, this method only gives -1 and 2 as the roots of the quadratic. But obviously this is not the correct answer as this inequality ( with equality ) even if x is put equal to -2.

What am I doing wrong , why am I obtaining the wrong interval ?

What is the correct method to solve inequalities of this kind ?

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    $\begingroup$ You can't always square inequalities. You should make sure that both sides are positive. $\endgroup$ – S.C.B. May 1 '16 at 5:50
  • $\begingroup$ Ahh yes , I had not considered that . Can you give me a hint on how to proceed while keeping this in mind ? $\endgroup$ – Noob101 May 1 '16 at 5:52
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HINT

You can't always square inequalities.

You should make sure that both sides are positive. And you also have to make sure the radicand is positive.

However, in this case, the inequality is not to hard to prove that it is true for $-2 \le x<0$, so it does not really matter. Now you merely have to check $x \ge 0$.

Now, you can square both sides and get $$x+2 \ge x^2 \Leftrightarrow (x+1)(x-2) \le 0$$

When faced with problems such as $$f(x) \ge g(x)$$Where both sides involve square roots, you should split it into two cases (Well, maybe not just two.) in order to avoid the above mentioned problem.

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Split the range into $3$ parts:

  • $\color\red{x<-2}\implies(x+2)^{\frac12}\not\in\mathbb{R}$

  • $\color\red{-2\leq x<0}\implies(x+2)^{\frac12}\geq0>x$

    So $\color\green{-2\leq x<0}$ is part of the solution.

  • $\color\red{x\geq0}\implies(x+2)^{\frac12},x\geq0$, therefore:

    You can square each side of the inequality without affecting the sign.

    As you've noted, solving the quadratic gives you $-1<x<2$.

    Intersecting this range with $x\geq0$ gives you $0\leq x<2$.

    So $\color\green{0\leq x<2}$ is another part of the solution.


Hence the combined solution is the union of:

  • $-2\leq x<0$
  • $0\leq x<2$

Or simply $-2\leq x<2$.

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You could also approach this by considering the properties of the curves for the functions $ \ y \ = \ x \ $ and $ \ y \ = \ \sqrt{x+2} \ $ . The square-root function only gives non-negative values and the curve $ \ y \ = \ \sqrt{x+2} \ $ is the curve for $ \ y \ = \ \sqrt{x} \ $ shifted horizontally "to the left" by 2 units; so the curve is always above the $ \ x- $ axis except at the point $ \ (-2, \ 0 ) \ $ . Thus, the curve plainly lies above the line $ \ y \ = \ x \ $ in the interval $ \ [ \ -2 \ , \ 0 \ ] \ $ .

The intersection point(s) of the two curves is found by setting the functions equal and solving for values of $ \ x \ $ (so it is all right to square both sides here):

$$ \ \sqrt{x + 2} \ = \ x \ \ \Rightarrow \ \ x^2 \ - \ x \ - \ 2 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ -1 \ \ , \ \ x \ = \ 2 \ \ , $$

as you've already found. The solution $ \ x \ = \ -1 \ $ is "spurious" since the square-root curve would have to meet $ \ y \ = \ x \ $ at $ \ ( -1, \ -1 ) \ $ , so there is only an intersection at $ \ (2, \ 2 ) \ $ . For $ \ x \ > \ 2 \ $ , the inequality is not satisfied: the line continues to rise above the square-root curve. So the inequality $ \ \sqrt{x + 2} \ > \ x \ $ is satisfied only on the interval $ \ [ \ -2 \ , 2 \ ) \ $ .

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