# Quadratic inequalities - why does the solution seem different for $(x-a)(x-b) <0$ and $(x-a)(x-b) >0$?

I want to ask about way of solving exponential inequalities, I am going to show you two similar examples, but their solving is kinda different.

First example: $$3^{2x}-10\cdot3^x+9>0$$ $$(3^x-9)(3^x-1)>0$$ $$3^x\in(-\infty;1)\cup(9;+\infty)$$ So to find x we need to solve:

$3^x>9$ and $3^x>1$ (1)

and we get: $$x\in(-\infty;0)\cup(2;+\infty)$$

Second example: $$5^{2x}-6\cdot5^x+5<0$$ $$(5^x-5)(5^x-1)<0$$ $$5^x\in(1;5)$$ So to find x we need to solve:

$\begin{cases}5^x<5\\5^x>1\end{cases}$ (2)

and we get: $$x\in(0;1)$$

My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)

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When you have $ab \gt 0$, you can have $a\gt 0, b \gt 0$ or $a \lt 0, b \lt 0$. So for your first example you should have $[3^x \gt 9$ and $3^x \gt 1]$ or $[3^x \lt 9$ and $3^x \lt 1]$. You can combine each of the square brackets because one of the inequalities is always true when the other one is to get $3^x \gt 9$ or $3^x \lt 1$ (note the or, not and), leading to the solution you have. For the second, if $ab \lt 0$ one of $a$ and $b$ is greater than zero and the other is less. So you can have $[5^x \gt 5$ and $5^x \lt 1]$ or $[5^x \lt 5$ and $5^x \gt 1]$. In this case the inequalities in the first bracket are inconsistent, so that one can be ignored and we get the inequalities you cite.

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There is no difference: in the first case you should solve $$(3^x>9\ \mbox{and}\ 3^x>1)\ \mbox{or}\ (3^x<9\ \mbox{and}\ 3^x<1),$$ and in the second case you should solve $$(5^x>5\ \mbox{and}\ 5^x<1)\ \mbox{or}\ (5^x<5\ \mbox{and}\ 5^x>1).$$ It just happens that $(3^x>9$ and $3^x>1)$ reduces to $(3^x>9)$, that $(3^x<9$ and $3^x<1)$ reduces to $(3^x<1)$, and that $(5^x>5$ and $5^x<1)$ is impossible.
@Didier: please consider the level of question; you're using logical notation that the OP probably hasn't encountered. Using $\lor$ and $\land$, and $<$ and $>$ in your explanation might therefore be quite confusing to sort out...? Couldn't you have simply written "or" and "and"? (No LaTeX required there!) –  amWhy May 4 '11 at 18:35
@Amy: Interesting statement of yours. Two points: (1) Do you seriously object to the use of $<$ and $>$ on MSE? (2) Please see math.stackexchange.com/questions/35368 –  Did May 4 '11 at 18:56
@Didier: I think it's more the use of $\land$ and $\lor$ as opposed to "and" and "or"; perhaps it is their proximity to $\lt$ and $\gt$ here that leads to their mention, rather than an objection to the latter per se. –  Arturo Magidin May 4 '11 at 19:03
@Didier: not at all (do I object to the use of $\lor$ nor to $\land$), when it's appropriate: clearly, the OP of the message you link to was immersed in studying logic! In this case, we can fairly well assume that the OP hasn't encountered much in the way of logic notation. I was suggesting that we need to gear answers so that they are understood by the questioners. If a question were asked, here, about how to solve a system of two equations in two variables: x + y = 4; 2x - y = 0 (e.g.) [tagged algebra/pre-cal] we should hardly answer by using a matrix! –  amWhy May 4 '11 at 19:08