please help with it:
square root(x+2-x^2)<square root(x^2-3x+2)+x
Square both sides:
$\displaystyle x^2+x+2 + x^2 -3x + 2 - 2(x-1)\sqrt{x^2 - 4} < x^2$
$\displaystyle x^2 - 2x + 4 < 2(x-1)\sqrt{x^2 - 4} $
$\displaystyle ((x-1)^2 + 3)^2 < 4(x-1)^2(x^2 - 4)$
Then it is ugly, very ugly.... I wonder if there is a neat solution.
Soroban is answering it anyway, so let us see...
The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
The problem printed there was: $\displaystyle \sqrt {x + 2 - x^2 } < \sqrt {x^2 - 3x + 2} + x$. Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.
$\displaystyle \sqrt {x + 2 - x^2 }$ is only defined in the interval $\displaystyle \left[ { - 1,2} \right]$.
Looking at the graphs of both sides, one can easily see that the only solution is $\displaystyle x \in [ - 1,0)$.
$\displaystyle \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$
$\displaystyle x^2 + x + 2 < (x^2 - 3x + 2) + x^2 + 2x\sqrt{x^2 - 3x + 2}$
$\displaystyle x^2 + x + 2 < 2x^2 - 3x + 2 + 2x\sqrt{x^2 - 3x + 2}$
$\displaystyle -x^2 + 4x < 2x\sqrt{x^2 - 3x + 2}$
Do NOT cancel that x! (If x is negative then we would be dividing by a negative number. Not kosher here as that changes the < to a >.)
$\displaystyle (-x^2 + 4x)^2 < 4x^2(x^2 - 3x + 2)$
$\displaystyle x^4 - 8x^3 + 16x^2 < 4x^4 - 12x^3 + 8x^2$
$\displaystyle -3x^4 + 4x^3 + 8x^2 < 0$
$\displaystyle 3x^4 - 4x^3 - 8x^2 > 0$
$\displaystyle x^2(3x^2 - 4x - 8) > 0$
$\displaystyle x^2 \left (x^2 - \frac{4}{3}x - \frac{8}{3} \right ) > 0$
The factoring here is a bit rough but can be done using the quadratic formula:
$\displaystyle x^2 \left ( x - \frac{1 + 2 \sqrt{7}}{3} \right ) \left ( x - \frac{1 - 2 \sqrt{7}}{3} \right ) > 0$
The critical points here are $\displaystyle x = 0, \frac{1 \pm 2 \sqrt{7}}{3}$, so we have four intervals on the real line to test in the original (that is to say, all the way back to the beginning!) inequality:
$\displaystyle x \in \left ( - \infty, \frac{1 - 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x$ (Nope!)
$\displaystyle x \in \left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$ (Check!)
$\displaystyle x \in \left ( 0, \frac{1 + 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x$ (Nope!)
$\displaystyle x \in \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$ (Check!)
So the solution set for x is
$\displaystyle \left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \cup \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right )$
(Note that there is a place where the domain of x is restricted. As this falls outside of the derived solution set I'm not going to worry over it.
(The graph is for the equivalent problem: $\displaystyle \sqrt{x^2 + x + 2} - ( \sqrt{x^2 - 3x + 2} + x) < 0$. We are looking for where the expression is negative.)
-Dan