# How can I solve the following square root inequality?

I am given the following inequality:

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

And I have to solve for $$x$$.

This is what I tried:

Firstly, I applied the existence condition for the square root:

$$2-x \ge 0 \Rightarrow x \le 2 \Rightarrow x \in (- \infty, 2]$$

Then I squared both sides of the inequality:

$$2-x>x^2$$

$$x^2+x-2 < 0$$

And from this I got that:

$$x \in (-2, 1)$$

Finally, I intersected this with the condition I applied at the beginning of the exercise:

$$x \in (-2, 1) \cap (- \infty, 2]$$

$$x \in (-2, 1)$$.

The problem with this answer is that it is wrong. If I take a number like $$-10$$ and plug it into the inequality I get:

$$\sqrt{2-(-10)} > -10$$

$$\sqrt{12} > -10$$

Which is true. However $$-10$$ is not included in the interval $$(-2, 1)$$. The correct answer seems to be $$(- \infty, 1)$$, which is not what I got.

I noticed that the inequalities of before and after of the squaring are not equivalent. So:

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

and

$$2-x>x^2$$

are not equivalent. If, again, I take the number $$-10$$, in the first inequality I get:

$$\sqrt{2-(-10)} > -10$$

$$\sqrt{12} > -10$$, which is true.

And in the second inequality I get:

$$2 - (-10) > (-10)^2$$

$$12 > 100$$, which is false.

So I think that's where the problem lies, but I don't know if I am correct and what should I do to get the right answer of $$(- \infty, 1)$$.

• When you're solving $$x^2 +x -2 > 0$$ you need to remember that previously you have found both $x\leq 2$ and $x>0$ (because square root cannot be negative). So in this case you're limited to $0 \leq x \leq 2$, before even looking at solving this equation. – Matti P. Oct 30 '19 at 13:29
• "Then I squared both sides of the inequality:" $x > y\not \implies x^2 > y$ unless $y \ge 0$. If $x > 0 > y$ then multiplying by $y$ flips the signs and $x>0 >y \implies xy < 0 < y^2$. And multiplying by $x$ doesn't flip and $x > 0 > y\implies x^2 > 0 >xy$ and we have $x^2 > xy$ and $xy < y^2$ and we can't conclude anything about $x^2$ compared to $y^2$. – fleablood Oct 31 '19 at 6:42
• @MattiP., the non-negativity restriction on the square root only implies $x\le2$ (as the OP found). It does not require $x\ge0$. Indeed, the inequality $\sqrt{2-x}\gt x$ is clearly true for all $x\lt0$. – Barry Cipra Oct 31 '19 at 12:45

"Then I squared both sides of the inequality:"

$$a > b\not \implies a^2 > b^2$$ unless $$b \ge 0$$.

If $$a > 0 > b$$ then multiplying by $$b$$ flips the signs and $$a>0 >b \implies ab < 0 < b^2$$. And multiplying by $$a$$ doesn't flip and $$a > 0 > b\implies a^2 > 0 >ab$$.

[Remember the reason that $$m > n > 0 \implies m^2 > n^2$$ is that because $$n > 0$$ we know $$m>n \implies mn > nn = n^2$$ and because $$m >0$$ we know $$m^2 = mm > mn$$. So we have $$m^2 >mn$$ and $$mn > n^2$$ so $$m^2 > n^2$$. That simply will not work of one of $$m,n$$ is negative. (and if both are negative you get the exact opposite result: $$0 > m>n\implies m^2 < n^2$$.]

So we have $$a^2 > ab$$ and $$ab < b^2$$ and we can't conclude anything about $$a^2$$ compared to $$b^2$$.

.... Or ....

just to be blunt.

$$3 > -987$$ by $$3^2 \not > (-987)^2$$.

....

But you can square both sides if you acknowedge you are assuming $$x \ge 0$$ as a conditional that may not be true.

so when you get $$x \in (-2,1)$$ IF $$x \ge 0$$ so $$[0,1)$$ you can intersect it with the positive values you had before. $$(-\infty, 2] \cap [0,1)=[0,1)$$ IFF $$x \ge 0$$.

But we CAN have $$x < 0$$ in which case ... squaring both sides is useless.

So either $$x < 0$$ and $$x\in (-\infty, 0)$$ OR $$x \ge 0$$ and $$x\in [0,1)$$ and so $$x \in (-\infty,0) \cup [0,1) = (-\infty, 1)$$>

The domain of the root is given by b$$x\le 2$$ Now we will consider two cases: If $$x\le 0$$ the our inequality is true. If $$x then we can square and we get $$0>x^2+x-2$$ This gives us $$-2 and $$0 and we get $$0 So the solution set is given by $$x<1$$ and $$x$$ is a real number.

$$\sqrt{2-x} > x$$ then $$\sqrt{2-x} > x-2 +2$$ , move all to the left hand side then $$2-x = (\sqrt{2-x})^2$$ $$\left(\sqrt{2-x}\right)^2 + \sqrt{2-x} - 2 > 0$$ $$( \sqrt{2-x} +2) (\sqrt{2-x} -1)>0$$ I avoided squaring both sides.

Starting from

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

it is immediately clear that if $$x<0$$ then $$\sqrt{2-x} >0$$ and therefore $$x<0$$ is part of the solution set. We then need to analyze where

$$\sqrt{2-x}$$

is defined. We know that $$\sqrt{2-x}$$ will have positive real solutions if $$2-x \ge 0$$. Therefore, we need to check $$0\le x \le 2$$. In order to do this, let $$\varepsilon$$ be a small number where $$0\le\varepsilon< 1$$. Then $$\sqrt{2-\varepsilon}>\sqrt{1}> \varepsilon$$ $$\sqrt{2-1}=\sqrt{1}=1$$ $$\sqrt{2-(\varepsilon+1)}\le\sqrt{1}\le\varepsilon+1$$

so when $$0\le x < 1$$ we have $$\sqrt{2-x} > x$$ and when $$1\le x \le 2$$ we have $$\sqrt{2-x} \le x$$. Hence, the proper solution set is $$x<1$$ or $$(-\infty,1)$$.

An alternative approach is to start from the fact that we must have $$x\le 2$$ in order for $$\sqrt{2-x}$$ to exist, and write $$x=2-u^2$$ with $$u\ge0$$, which simplifies the inequality to $$u\gt2-u^2$$ (with, remember, $$u\ge0$$, which is used in removing the square root symbol to obtain $$\sqrt{u^2}=u$$). Standard algebra manipulates this to the equivalent form $$(u-1)(u+2)\gt0$$, which, given the restriction to $$u\ge0$$, implies $$u\gt1$$. Substituting back to $$x=2-u^2$$, we see we have $$x\lt2-1^2=1$$, so the solution set for the inequality is $$x\in(-\infty,1)$$.

In review, you have to do something to get rid of the square root symbol in order to find the solution set for the inequality. Squaring both sides certainly does this, but you have to contend with the fact that $$a\gt b$$ is not equivalent to $$a^2\gt b^2$$ without additional restrictions on $$a$$ and $$b$$, so that after you've found the solution set for the squared inequality you have to see how it relates to the original inquality. The approach here is somewhat simpler in that regard; in essence, it puts all its restrictive eggs into a single basket of non-negativity (i.e., $$u\ge0$$).

I think fleablood’s answer gets closest to addressing what I think OP was asking, namely why the mechanical application of squaring both sides of the equation does not yield all of the solutions.