Need the domain of integral values that satisfy

[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 $$5 \le x \le 35$$?

 

[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.

 

[cscott]

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.

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

 

[assyrian_77]

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

 

[cscott]

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

The second one can't be right...

 

[assyrian_77]

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

Correct?

Yep. And what can you say from this?

 

[cscott]

Well, the first would be redundant.

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

 

[HallsofIvy]

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

The second one can't be right...

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.

 

[VietDao29]

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. I can say $$5 \le x \le 35$$, but this doesn't hold true for the inequality with 900 on the RHS because the values go below zero for a bit (which doesn't make sense in this problem). What am I missing here?

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? :)

 

[cscott]

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

 

[VietDao29]

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

Yes, it's correct.