Can You Solve These Rational Equations?

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The discussion focuses on solving two rational equations: i) (x+6)/(x+2) ≤ x and k) (4x+4)/(x-1) < x^2 + 3x + 2. For equation i), the solution is found to be x in the intervals [-3, -2] and (2, ∞). In equation k), there is confusion regarding the correct form of the numerator, which should be clarified as 4x + 4. Participants emphasize the importance of using LaTeX for clarity in mathematical expressions and suggest revising the setup of the sign chart for accurate critical points. The discussion concludes with a participant expressing gratitude for the assistance received in solving the problem.
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Without actually giving you all the details of the process for #i,

Subtract x from both sides;
Multiply both sides by (x+2);

The left-hand member can be expressed as x2+x-6.
You need to consider two cases. (x+2) positive, and (x+2) negative. Also, as you might have noticed, x cannot be -2.
 
anonymous12 said:
k) 4x - 4/ x - 1 < x^2 + 3x + 2
Need parentheses! This looks like
4x - \frac{4}{x} - 1 &lt; x^2 + 3x + 2
Furthermore, looking at your work, it looks like the the numerator was originally 4x + 4. So which is it? 4x - 4 or 4x + 4? Please double check the problem.
 
eumyang said:
Need parentheses! This looks like
4x - \frac{4}{x} - 1 &lt; x^2 + 3x + 2
Furthermore, looking at your work, it looks like the the numerator was originally 4x + 4. So which is it? 4x - 4 or 4x + 4? Please double check the problem.

Sorry, the question is (4x+4)/(x-1) < x^2 + 3x + 2 and by the way how did you write the problem as:

4x - \frac{4}{x} - 1 &lt; x^2 + 3x + 2
 
anonymous12 said:
Sorry, the question is (4x+4)/(x-1) < x^2 + 3x + 2 and by the way how did you write the problem as:

4x - \frac{4}{x} - 1 &lt; x^2 + 3x + 2

There's a button in the editing window (the last one on the 2nd row, the one that looks like the summation symbol) that allows you to type in LaTeX. You should learn it -- makes the expressions/equations/inequalities much more readable.

Anyway, I think your problem here is that you set up the sign chart too early. You have this:
\frac{(x+1)(x+2)(x-1)\text{ }(-4x-4)}{x-1} &lt; 0
... and the problem is that it looks like the last factor (-4x-4) is multiplied by the other 3 factors, when in fact it is added (or subtracted). And so you didn't get the correct zeros for the numerator when looking for the critical points.

What you'll need to do is take the numerator (which should be written like this):
(x+1)(x+2)(x-1) - 4x - 4
multiply out the 3 factors, combine like terms with the -4x - 4, and THEN set equal to zero. Since this will be a cubic, you'll have to try the Rational Roots test. (The 3 roots are all integers, which makes things a little easier.)
 
OMG! Thank you so much! This question was driving me crazy and thanks to you I finally solved it! :)
 
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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