Solve 1/x < 5+1/(x^2+x) ???HELP

[henribar]

The interval(s) where 1/x is less than 5+1/(x^2+x) is not the same interval(s) where 0 is less than (5x^3+4x^2)/(x^3+x^2) which is just 1/x < 5+1/(x^2+x) rearranged . This makes absolutely no sense I've double, triple, quadruple checked my answers even using graphing software. Please help i'm losing my mind.

 

[Deleted member 4993]

The interval(s) where 1/x is less than 5+1/(x^2+x) is not the same interval(s) where 0 is less than (5x^3+4x^2)/(x^3+x^2) which is just 1/x < 5+1/(x^2+x) rearranged . This makes absolutely no sense I've double, triple, quadruple checked my answers even using graphing software. Please help i'm losing my mind.

5 + 1/(x² + x) = 5 + 1/x - 1/(x+1)

continue....

 

[stapel]

\(\displaystyle \mbox{Solve }\, \dfrac{1}{x}\, <\, 5\, +\, \dfrac{1}{x^2\, +\, x}\)

The interval(s) where 1/x is less than 5+1/(x^2+x) is not the same interval(s) where 0 is less than (5x^3+4x^2)/(x^3+x^2) which is just 1/x < 5+1/(x^2+x) rearranged.

Since the LCM of "x" and "x^2 + x = x(x + 1)" is x(x + 1), I'm not seeing how you arrived at something with a denominator of x^3 + x^2...? Also, how did you arrive at your conclusion regarding the interval(s)? Which interval(s) did you get, anyway?

Please reply showing your steps and reasoning. You started with:

. . . . .\(\displaystyle \dfrac{1}{x}\, <\, 5\, +\, \dfrac{1}{x^2\, +\, x}\)

. . . . .\(\displaystyle \dfrac{1\, (x\, +\, 1)}{x\, (x\, +\, 1)}\, <\, \dfrac{5x\, (x\, +\, 1)}{x\, (x\, +\, 1)}\, +\, \dfrac{1}{x\, (x\, +\, 1)}\)

. . . . .\(\displaystyle \dfrac{x\, +\, 1}{x\, (x\, +\, 1)}\, <\, \dfrac{5x^2\, +\, 5x}{x\, (x\, +\, 1)}\, +\, \dfrac{1}{x\, (x\, +\, 1)}\)

. . . . .\(\displaystyle \dfrac{x\, +\, 1}{x\, (x\, +\, 1)}\, <\, \dfrac{5x^2\, +\, 5x\, +\, 1}{x\, (x\, +\, 1)}\)

And... then what? Please be complete. Thank you!

 

[henribar]

I made a typo, 1/x<5+(1/(x^2+x)) can be rearranged to 0<(5x^2+4x)/x(x+1) NOT (5x^3+4x^2)/(x^3+x^2) . My question still stands, however.
the intervals where 1/x is less than 1/(x^2+x) should be the SAME intervals where 0 is less than (5x^2+4x)/x(x+1) but they aren't

Answer to your question (to get to 0<(5x^2+4x)/x(x+1):
From:
x+1x(x+1)<5x2+5x+1x(x+1)

bring (x+1)/(x(x+1)) to the right side of the equation by subtracting it on both sides, then (5x^2+5x+1-x-1)/x(x+1) since denominators are the same
and voila.
0<(5x^2+4x)/x(x+1)
So whats the deal??
betw

 

[henribar]

As far as getting the intervals i'm not quite sure how I would show my work on here but simply typing in the original inequality 1x<5+1x2+x


and seeing the intervals you can see how they differ from the intervals of my the rearranged inequality. I used Desmos

 

[henribar]

I solved my problem, yes, 1x<5+1x2+x c
cannot be rearranged to
5x^3+4x^2)/(x^3+x^2) , that was a math error on my part. The question still stood however when rearranged to (5x+4)/(x+1); what I figured out was I had a fundamental misunderstanding when it came to reading and understanding inequalities. Hard to explain on here, but I am happy to report there is not a glitch in mathematics, and I have successfully regained my sanity.