# Need the domain of integral values that satisfy

1. Mar 13, 2006

### cscott

I have this inequality:

$$4x^2 - 160x + 1500 \le 900$$

I brought the 900 to the LHS and found the roots (35.81, 4.19). Now I just need the domain of integral values that satisfy. Can I say $$5 \le x \le 35$$?

Last edited: Mar 14, 2006
2. Mar 13, 2006

### assyrian_77

The range ($5\le x\le 35$) you have is not correct. Thinking of like this might help get you a range: $(x-35.81)(x-4.91)\le 0$. From the solution you can see what range you should have.

3. Mar 13, 2006

### cscott

Sorry, I meant domain not range. Or is that what you meant too?

4. Mar 13, 2006

### assyrian_77

$(x-35.81)(x-4.91)\le 0$ gives you two inequalities, right? What can you conclude from them?

5. Mar 13, 2006

### cscott

$$x \le 35.81$$
$$x \le 4.91$$

The second one can't be right...

6. Mar 13, 2006

### assyrian_77

Yep. And what can you say from this?

7. Mar 13, 2006

### cscott

Well, the first would be redundant.

How can this be correct, because if I substitute x = 3 in my orginal inequality, I get $1056 \le 900$

8. Mar 14, 2006

### HallsofIvy

No, that's not right. Think about the graph of a quadratic. If
$$y= 4x^2 - 160x + 1500$$ that's a parabola opening upward.
Where is
$$y= 4x^2 - 160x + 1500= 0$$
y will be negative between them.

9. Mar 14, 2006

### VietDao29

Hmm, okay, say you have the a quadratic function:
f(x) := ax2 + bx + c, which has 2 solutions:
x1, and x2. (And x1 < x2)
Then if x0 is in ]x1, x2[
Then af(x0) < 0, that means if a > 0, then f(x0) < 0, and vice versa, if a < 0, then f(x0) > 0.
If x0 is not in ]x1, x2[, then af(x0) > 0.
---------------
Example:
Solve:
x2 - 3x - 1 > -3
<=> x2 - 3x + 2 > 0
<=> (x - 2) (x - 1) > 0
<=> x < 1, or x > 2 (since a = 1 > 0, x1 = 1, and x2 = 2).
So can you apply it to your problem? :)

Last edited: Mar 14, 2006
10. Mar 14, 2006

### cscott

So if I'm interpreting you guys correctly, then the domain in my OP is correct? (5 <= x <= 35)

11. Mar 16, 2006

### VietDao29

Yes, it's correct.