# Math Help - root

1. ## root

square root(x+2-x^2)<square root(x^2-3x+2)+x

2. Originally Posted by agripa

square root(x+2+x^2)<square root(x^2-3x+2)+x
Square both sides:

$x^2+x+2 + x^2 -3x + 2 - 2(x-1)\sqrt{x^2 - 4} < x^2$
$x^2 - 2x + 4 < 2(x-1)\sqrt{x^2 - 4}$
$((x-1)^2 + 3)^2 < 4(x-1)^2(x^2 - 4)$

Then it is ugly, very ugly.... I wonder if there is a neat solution.
Soroban is answering it anyway, so let us see...

3. Originally Posted by agripa
square root(x+2+x^2)<square root(x^2-3x+2)+x
The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
The problem printed there was: $\sqrt {x + 2 - x^2 } < \sqrt {x^2 - 3x + 2} + x$. Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.

$\sqrt {x + 2 - x^2 }$ is only defined in the interval $\left[ { - 1,2} \right]$.
Looking at the graphs of both sides, one can easily see that the only solution is $x \in [ - 1,0)$.

4. Originally Posted by Plato
The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
The problem printed there was: $\sqrt {x + 2 - x^2 } < \sqrt {x^2 - 3x + 2} + x$. Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.

$\sqrt {x + 2 - x^2 }$ is only defined in the interval $\left[ { - 1,2} \right]$.
Looking at the graphs of both sides, one can easily see that the only solution is $x \in [ - 1,0)$.
But lets say this mistaken question is posed, then do you have a solution?

Thanks

5. Originally Posted by agripa

square root(x+2+x^2)<square root(x^2-3x+2)+x
$\sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$

$x^2 + x + 2 < (x^2 - 3x + 2) + x^2 + 2x\sqrt{x^2 - 3x + 2}$

$x^2 + x + 2 < 2x^2 - 3x + 2 + 2x\sqrt{x^2 - 3x + 2}$

$-x^2 + 4x < 2x\sqrt{x^2 - 3x + 2}$

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 >.)

$(-x^2 + 4x)^2 < 4x^2(x^2 - 3x + 2)$

$x^4 - 8x^3 + 16x^2 < 4x^4 - 12x^3 + 8x^2$

$-3x^4 + 4x^3 + 8x^2 < 0$

$3x^4 - 4x^3 - 8x^2 > 0$

$x^2(3x^2 - 4x - 8) > 0$

$x^2 \left (x^2 - \frac{4}{3}x - \frac{8}{3} \right ) > 0$

The factoring here is a bit rough but can be done using the quadratic formula:
$x^2 \left ( x - \frac{1 + 2 \sqrt{7}}{3} \right ) \left ( x - \frac{1 - 2 \sqrt{7}}{3} \right ) > 0$

The critical points here are $x = 0, \frac{1 \pm 2 \sqrt{7}}{3}$, 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:
$x \in \left ( - \infty, \frac{1 - 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x$ (Nope!)

$x \in \left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$ (Check!)

$x \in \left ( 0, \frac{1 + 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x$ (Nope!)

$x \in \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x$ (Check!)

So the solution set for x is
$\left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \cup \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right )$

(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: $\sqrt{x^2 + x + 2} - ( \sqrt{x^2 - 3x + 2} + x) < 0$. We are looking for where the expression is negative.)

-Dan