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Need to solve the following and write the answer in interval notation.

$x^2-6x+9<0$

I know that factors to $(x-3)^2<0$. I presume I need to take the sqrt of both sides but I'm stuck from that point.

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Do case analysis for two different cases (I am going to use the example you have given).

Case 1: $(x-3)$ is positive. Then $x-3 \lt 0$ (Divide both sides by $x-3$). But wait! $x-3$ is said to be positive, which means $x-3 \gt 0$. Then $x-3$ can't be positive. That means no solution for case 1.

Case 2: $(x-3)$ is negative. Then $x-3 \gt 0$ (Divide both sides by $x-3$). But $x-3$ is said to be negative, so no solution for this case too. $x-3 = 0$ also won't work since $0$ is not less than 0.

This happens to be a example where there is no solution. In this particular case, this can be achieved faster: set $x-3 = u$. Then $u^2 \lt 0$. Since squaring a real number can never be negative, there's no solution.

The general approach to these kinds of problem is as follows: separate the problem into cases and analyze those case by case.

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  • $\begingroup$ So you're saying that $(x-3)^2 < 0$ equates to $x-3<0$ and $x-3>0$ can you explain that part? I understand the rest. If i were to take the sqrt of both sides it would be +- sqrt(x-3) $\endgroup$ – dstarh Oct 23 '18 at 2:09
  • $\begingroup$ Two "return" characters will separate paragraphs in the body of a Question or an Answer (but not in a Comment). $\endgroup$ – hardmath Oct 23 '18 at 2:14
  • $\begingroup$ I am sorry for the delay. Yes, I will explain. These are 3 scenarios to this inequality. One which is x-3 is greater than 0. Other is x-3 is smaller than 0. Also, x-3 could be equal to 0. These two are two possible scenarios to this inequality (x-3)^2 < 0. (x−3)^2<0 does not equates to x−3<0 and x−3>0, rather, it has three possible scenarios. It is important to set those scenarios is because dividing a negative number changes the 'direction' of the inequality (ex. changing < to >) while dividing a positive number doesn't. You can think of taking the square root as dividing by x-3. (not finish) $\endgroup$ – hengyi sun Oct 23 '18 at 2:44
  • $\begingroup$ (continue) Then, it should be easier to see why dividing into cases is necessary. Dividing by a positive number doesn't change the direction of the inequality, dividing by a negative number does change it. And dividing by 0? Well, for that, one could just set whatever equal to 0, 0. Since all three might lead to different outcome, it is important to do it in separate cases. I hope that helps you. $\endgroup$ – hengyi sun Oct 23 '18 at 3:01
  • $\begingroup$ @hardmath Thanks you very much $\endgroup$ – hengyi sun Oct 23 '18 at 3:02

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