Pre-calculus; linear inequalities

AI Thread Summary
The discussion focuses on solving two quadratic inequalities: 9x^2 + 12x + 4 ≥ 0 and 2 - 5x - 3x^2 ≤ 0. The first inequality can be factored as (3x + 2)(3x + 2), leading to conclusions about the values of x that satisfy it. The second inequality was incorrectly transformed into a non-inequality form, prompting clarification on the correct expression and the need to identify sign errors. Additionally, participants noted that these problems are quadratic inequalities, not linear inequalities as initially stated.
priscilla98
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Homework Statement



1. 9x^2 + 12x + 4 ≥ 0

2. 2 - 5x - 3x^2 ≤ 0

Homework Equations



The Attempt at a Solution



1. Can't you factor this into (3x + 2) (3x + 2)?
2. Looking at this inequality, I am thinking if you can change this inequality into 3x^2 + 5x + 2.
 
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priscilla98 said:

Homework Statement



1. 9x^2 + 12x + 4 ≥ 0

2. 2 - 5x - 3x^2 ≤ 0

Homework Equations



The Attempt at a Solution



1. Can't you factor this into (3x + 2) (3x + 2)?

Yes, you can. So what is your conclusion about the x that satisfy it?

2. Looking at this inequality, I am thinking if you can change this inequality into 3x^2 + 5x + 2.

What you have changed it into isn't an inequality and you likely have a sign error. Did you mean to write 3x2 + 5x - 2 ≥ 0? If so, what next?

Also, your title says linear inequalities. These aren't linear inequalities; they are quadratic inequalities.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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