Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did wrong:
$-1
Possible solutions:
$2
$2
$3
$3
Therefore $S=\{2
The only problem is that the correct solution is $S=\{1
Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did wrong:
$-1
Possible solutions:
$2
$2
$3
$3
Therefore $S=\{2
The only problem is that the correct solution is $S=\{1
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)
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.