Solving $a^2-3a+2<0$. Okay. I am confused in this part. 
$$(a-1)(a-2)<0$$ $$a<1;a<2$$ So $$a\in(-\infty,1)$$. But the answer is $$a\in(1,2)$$ How please help? 
Edit: I still cannot understand why we take the latter to be the correct answer. But if I ask you what will be the solutions of $$(a-1)(a-2)(a-4)<0$$ then? 
Edit: Okay, I understood. Because the product of the two numbers is negative! But how will you solve the previous $3$ term inequality?
 A: Look at the algebraic sign of $(a-1)(a-2)$ for different values of $a$, remembering how the sign of the product of two numbers behaves:
       a-1:       -         0      +     +         +  
       a-2:       -         -      -     0         +  
(a-1)(a-2):       +         0      -     0         +
           -----------------1------------2-------------------  
                      number line for value of a

As you can see from the diagram, $(a-1)(a-2)$ is a product of two negative numbers and is therefore positive when $a<1$. When $a>2$ it’s a product of two positive numbers and is again positive. Of course it’s $0$ when one of the factors is $0$, but when $a$ is between $1$ and $2$, the $a-1$ factor is positive, and the $a-2$ factor is negative, so the product is negative.
Added. You can use the same idea for a product of any number of linear factors. For the inequality $(a-1)(a-2)(a-4)<0$ the diagram becomes:
       a-1:       -         0  +  +       +       +  +
       a-2:       -         -  -  0       +       +  +
       a-4:       -         -  -  -       -       0  +
(a-1)(a-2):       -         0  +  0       -       0  +  
           -----------------1-----2---------------4----  
                      number line for value of a

Now all three factors are negative when $a<1$, so their product is negative. When $a$ is between $1$ and $2$, two of them are negative and the third is positive, so their product is positive. When $a$ is between $2$ and $4$, only one factor is negative, so their producis negative. Finally, when $a>4$ all three factors are positive, and so is their product. Thus, the solution set is the union of two intervals: $(-\infty,1)\cup(2,4)$.
A: Step 1: First solve $a^2-3a+2=0$ which gives $a=1$ and $a=2$
Step 2: Put the solutions on a number line. Now you have three regions. For each region, pick a number and put that in the inequality. So for example, pick $a=-10$ which gives you a positive outcome. Pick $a=1.5$ which gives you a negative outcome. Pick $a=10$ which gives you a positive outcome. So the signs on the numberline are$+ , - , +$ Step 3: Since the inequality is $<0$, it follows that $a$ needs to be between $1$ and $2$
A: Unfortunately your first line of working is spurious: $a<1$;$a<2$ which other answers have shown does not hold. I think solving this graphically would be the best way to see it. You have correctly factorised to $(a-1)(a-2) = 0$ so you can find the zeroes of the function i.e. the x-intercepts of the parabola. 
A: 
Theorem. For all real numbers $x$ and $y$, we have $xy < 0$ iff ($x<0$ and $y>0$) OR ($x>0$ and $y<0$).

So $(a−1)(a−2)<0$ iff


*

*$a-1<0$ and $a-2>0$, or

*$a-1>0$ and $a-2<0$.


So $(a−1)(a−2)<0$ iff


*

*$a<1$ and $a>2$, or

*$a>1$ and $a<2$.


So $(a−1)(a−2)<0$ iff


*

*$a>1$ and $a<2$.

A: $(a−1)(a−2)<0$
$\implies \{a<1$ and $a>2\}$ or $\{a>1$ and $a<2\}$
You can see that there is no real number satisfing $\{a<1$ and $a>2\}$, so the answer is $1<a<2$.
A: Well first of all you are wrong as far as your hypothesis, which should be (-inf,2). Anyway the two parenthesis cannot be both less than 0. Find out why is this you have kl that k,l< 0 that means kl>0. Then as far as 2>1 this means that when (a-1)<0=>a<1=>(a-2)<0 so a must be in (1,2)
