6
$\begingroup$

Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did wrong:

$-1 $-6 $-6+\frac{25}{4} $\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$ $\frac{\pm\sqrt1}{\sqrt4}<\sqrt{\left(x-\frac{5}{2}\right)^2}<\frac{\pm\sqrt9}{\sqrt4}$ $\frac{\pm1}{2} $\frac{5\pm1}{2}

Possible solutions:

$2 (not valid)

$2 (ok)

$3 (not valid)

$3 (ok, but is a subset of solution 2.

Therefore $S=\{2

The only problem is that the correct solution is $S=\{1. Where am I wrong?

  • 0
    @LuizBorges The correct equivalence is \frac{1}{4}<\left( x-\frac{5}{2}\right) ^{2}<\frac{9}{4}\Leftrightarrow \frac{1}{2}<\left\vert x-\frac{5}{2}\right\vert <\frac{3}{2}.2012-05-03

2 Answers 2

3

You have a $\pm$ on each side of the inequality, but you need to change the direction of inequality for the "minus".

So you would have $\dfrac{5 + 1}{2} < x < \dfrac{5+3}{2}$ (The $+$'s go together), or $\dfrac{5 - 1}{2} > x > \dfrac{5 - 3}{2}$ (The $-$'s go together)

  • 0
    Hum... I think I understand it better now. I'm not at home now, so tomorrow I will sit with pencil and paper and try it out separating the two equations to visualize what you're saying, but I guess I'm almost there. Square roots and absolute values are much worser and counterintuive than I thought.2012-05-04
6

When you take the root of an inequality, you have to make sure that everything is positive and then take positive roots.

So after taking the roots, you get:

$\frac 12 < |x-\frac 52| < \frac 32$.

Now, you can regard the two cases $x> \frac 52$ and $x < \frac 52$ to eliminate the absolute value.

  • 0
    (For Luiz) - When x > \dfrac{5}{2}, the absolute value can be dropped. This gives the first inequality in my answer. When x < \dfrac{5}{2}, you must multiply everything by $-1$, giving the second inequality in my answer.2012-05-03