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If $x^2-2<frac{7}{2}x$ then what is the solution set for $x$?

I have most of the problem done, I just don't know how to lay out my answer. The answer is supposed to be $$boxed{-frac12<x<4}$$

First, I rewrote the problem saying $x^2-frac{7}{2}x-2<0$, and to be able to conveniently factor it I decided to multiply both sides by 2:

$$2x^2-7x-4<0$$

$$(2x+1)(x-4)<0$$

$$x<-frac12, x<4, text{or} (-infty, -frac12) cup (-infty, 4)$$

I know that doesn't make sense because that's the solution set for the equation less than $0$, when really we want to find the solution set for it less than $frac72x$. So how do I get the above solution? Would it suffice to just draw a number line and say if all values less than $-frac12$ and all values greater than $4$ turn the equation to $<0$ then it must be values between $-frac12 text{and} 4$?

You are on the right track. From $(2x+1)(x-4)<0$, you know that equality occurs when $x=-1/2$ and when $x=4$. To be safe, check the intervals begin{align}x<-1/2:quad &2x+1<0,quad x-4<0implies(2x+1)(x-4)>0\-1/2<x<4:quad&2x+1>0,quad x-4<0implies(2x+1)(x-4)<0\x>4:quad&2x+1>0,quad x-4>0implies(2x+1)(x-4)>0end{align} Therefore the only solution to the inequality is $-1/2<x<4$.

For $(2x+1)(x-4)<0$ to be true, you want one of the numbers $(2x+1), (x-4)$ to be negative, and the other to be positive.

Option 1: the first is negative:

$2x+1<0, x-4>0$ translates to $x<-frac12$ and $x>4$ which is, of course, impossible.

Option 2: The first is positive:
$2x+1>0, x-4<0$ translates to $x>-frac12$ and $x<4$ which means $xin(-frac12, 4)$.

The inequality is $(x-frac 7 4)^{2}<2+frac {49} {16}$. This means $-r <x-frac 7 4 <r$ where $r=sqrt {2+frac {49} {16}}=frac 9 4$. Can you continue?

The key realization is that $ab<0$ if and only if one of them is positive, and one of them is negative.

So we know that

$$2x+1 <0 implies x < - frac 1 2 ;;;;; text{and} ;;;;; x-4<0 implies x<4$$

But notice that, for some values of $x$, you might have some of these positive, or at other times negative. So we have to consider three potential solution sets:

• 
  $x$ is between $-1/2,4$
  


• 
  $x$ is less than $-1/2$
  


• 
  $x$ is greater than $4$
  

If $x<-1/2$, then $2x+1<0$, but $x-4<0$ too, so their product is positive.

Similarly, if $x>4$, then both factors and thus their product is positive.

Thus, neither solution set can satisfy the inequalities simultaneously (and thus cannot solve the original inequality). The only conclusion left is that $-1/2 < x < 4$.

Indeed, since $x>-1/2$, we have $2x+1>0$, but since $x<4$ we have $x-4<0$. This means their product is negative, which is precisely what we want!

A more intuitive method might be to consider that, as noted, the original inequality is

$$x^2 - frac 7 2 x - 2 < 0$$

When is a quadratic negative? As it happens, you can easily see, if the leading coefficient is positive, then the quadratic is negative between its two roots! (For negative leading coefficient, it is negative outside of its roots instead.)

Yet another method might be to consider the three cases for this inequality: either $x$ is to the left of the leftmost root, in between the two, or to the right of the rightmost root. Plug in some $x$ on each interval - say, $x=-1, x=0, x=5$, respectively - and see which give you cases less than zero, and those that don't. Owing to how the quadratic behaves, the corresponding "test values" that satisfy the inequality are on the interval that is the solution (left of the left root, in-between, or right of the right root), and thus your answer would be those intervals. (Granted this isn't much different than the method noted in the previous paragraph, but if you find this easier to process more power to you man.)

And of course, a graph always helps if it's available. Though a graph doesn't constitute a full-blown proof, it can always be helpful for getting an intuition of the situation. Indeed for the inequality above, we have the graph below, which makes the solution set immediately clear: