Problem:

Find all $x$ such that $|x^2-3x+1|<1$

I can't understand how to get started with this. I've never tried to solve quadratic Inequalities before. At first I thought of working with the Definition of the Modulus Sign, but then thought it would become too complicated. Any help with this problem would be greatly appreciated. Many thanks! 

• Does my answer help? – Prasun Biswas Jun 29 '15 at 7:29
• @PrasunBiswas Yes it does! Thanks a lot! – Ishan Jun 29 '15 at 7:31

$$|x^2-3x+1|\lt 1\iff -1\lt x^2-3x+1\lt 1$$

We have, using the lower bound,

\begin{align}-1\lt x^2-3x+1\iff x^2-3x+2\gt 0&\iff (x-2)(x-1)\gt 0\\&\iff x\in (-\infty,1)\cup (2,\infty)\end{align}

Also, we have, using the upper bound,

\begin{align}x^2-3x+1\lt 1\iff x(x-3)\lt 0\iff x\in (0,3)\end{align}

For both upper and lower bounds to hold, we need the intersection of these solution sets which is $(0,1)\cup (2,3)$.

Inequalities used:

• $ab\lt 0\iff a\gt 0~\land~b\lt 0\quad\lor\quad a\lt 0~\land~b\gt 0$

• $ab\gt 0\iff a\gt 0~\land~b\gt 0\quad\lor\quad a\lt 0~\land~b\lt 0$

You can use the inequality: $|A| < |B| \iff A^2 < B^2 \iff (A-B)(A+B) < 0$. Let $A = x^2-3x+1, B = 1 \Rightarrow (x^2-3x+1-1)(x^2-3x+1+1) < 0 \Rightarrow (x^2-3x)(x^2-3x+2) < 0 \Rightarrow x(x-3)(x-1)(x-2) < 0$. Can you finish it? Take a number $x = 4$, and plug it into the left side we see that the left side $> 0$, thus the solution is: $(0,1)\cup (2,3)$

• Sir, I made the edit. Sorry for the mistake. Sir, please could you elaborate your method a bit? I'm afraid I couldn't quite understand. – Ishan Jun 29 '15 at 7:11
• See me edit. Can you complete the "solution"? – DeepSea Jun 29 '15 at 7:17
• Yes Sir. Sir, could you please do me a favour? Sir, please could you have a look at my attempted solution and tell me where I've gone wrong? – Ishan Jun 29 '15 at 7:24
• Sir, I just understood where I was going wrong. Many thanks for your help Sir! – Ishan Jun 29 '15 at 7:30