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 $(- \infty, 1)$.
Best Answer
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 $$x<x\le 2$$ then we can square and we get $$0>x^2+x-2$$ This gives us $$-2<x<1$$ and $$0<x<1$$ and we get $$0<x<1$$ So the solution set is given by $$x<1$$ and $x$ is a real number.