Solve

- mathmath333

Solve

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- mathmath333

\(\large \color{black}{\begin{align} 2x^2+\dfrac{1}{x}\gt0\hspace{.33em}\\~\\
\end{align}}\)

- Sepeario

x>0 is a solution.

- MrNood

multiply both side od inequality by x
subtract 1 from both sides
divide both sides by 2
take cube root

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## More answers

- freckles

@MrNood if we multiply both sides by x. We have two cases. x is positive and x is negative.

- freckles

if we assume x is positive the inequality sign's direction holds
if we assume x is negative we flip the inequality sign's direction

- freckles

if you want to avoid that can you try multiplying x^2 on both sides instead

- mathmath333

how to solve this
\(x^3+\dfrac12>0\)

- freckles

subtracting 1/2 on both sides
then taking cube root of both sides

- mathmath333

u mean this ?
\(\large \color{black}{\begin{align} x>\left(-\dfrac12\right)^{1/3}\hspace{.33em}\\~\\
\end{align}}\)
x can only have real values

- freckles

that is right

- freckles

and since ()^(1/3) is an odd function you can actually bring the negative outside if you prefer

- freckles

\[x>- \frac{1}{2^\frac{1}{3}}\]

- mathmath333

but wolfram gives it as complex number
http://www.wolframalpha.com/input/?i=%5Cleft%28-%5Cdfrac12%5Cright%29%5E%7B1%2F3%7D%3D

- freckles

yes there are complex cube roots of -1/2
but you want the real cube root of -1/2

- freckles

http://www.wolframalpha.com/input/?i=real+cube+root+of+%28-1%2F2%29

- mathmath333

can this be done in case of square root too ?

- freckles

what do you mean exactly ?

- mathmath333

this one
http://www.wolframalpha.com/input/?i=real+square+root+of+%28-1%2F2%29

- freckles

short answer: no there is no real square root of -1/2
here is a long answer involving demorive' theorem
\[\frac{-1}{2}=\frac{1}{2}(\cos(\pi+2\pi n)+i \sin(\pi+2 \pi n) ) \\ \frac{-1}{2}=\frac{1}{2}e^{(\pi+2n \pi)i} \\ \text{ taking \square \root of both sides } \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2}(\pi+2n \pi)i} \text{ note: just raised both sides \to } \frac{1}{2} \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2} (\pi i)} = (\frac{1}{2})^\frac{1}{2}(\cos(\frac{\pi}{2})+i \sin(\frac{\pi}{2})) =(\frac{1}{2})^\frac{1}{2}(0+i)=(\frac{1}{2})^\frac{1}{2} i \\ \sqrt{\frac{-1}{2}}=(\frac{1}{2})^\frac{1}{2}e^{\frac{1}{2}(\pi+2 \pi)}=(\frac{1}{2})^\frac{1}{2}(\cos(\frac{3\pi}{2})+i \sin(\frac{3 \pi}{2}))=(\frac{1}{2})^\frac{1}{2}(-1i)\]
in other words square root of a negative number is only going to lead to imaginary numbers

- freckles

though that was an answer involving trig
I think I can also give you a good algebraic reasoning that has nothing to do with trig
the sqrt(x) function looks like this:
|dw:1436970600020:dw|
this doesn't exist to the left of the y-axis
so sqrt(-1) is not going to have a real answer
but
cuberoot(x) function looks like this:
|dw:1436970632089:dw|

- freckles

the cube function actually does exist to the left of the y-axis

- mathmath333

u mean even roots dont have real roots

- freckles

I think I said square roots

- freckles

yes i did
that is what I meant

- freckles

square root of negative numbers aren't real

- mathmath333

i mean if \(\Large a^{1/b}\) if b is even then there is no real root ?

- freckles

http://www.wolframalpha.com/input/?i=4th+roots+of+%28-1%2F2%29
though if you get into higher roots
we can also use demorive' theorem to show this
but I actually think you are only suppose to consider real function and real roots

- MrNood

all even roots are roots of the square root - so if the squre root is not real neither is any even root

- freckles

http://www.wolframalpha.com/input/?i=6th+roots+of+%28-1%2F2%29
yes even roots of negative numbers aren't going to have real
and the word even means also two which we call the square root

- freckles

the example I showed though was of the square root of a negative number since you asked about sqrt(-1/2)

- mathmath333

ok next qustion
\(\large \color{black}{\begin{align} \dfrac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1\hspace{.33em}\\~\\
\end{align}}\)

- freckles

I would probably end up subtracting 1 on both sides
and combining the fractions
but that seems really long

- freckles

only because there is so much multiplication we have to do in the numerator

- freckles

\[\frac{(x-1)(x-2)(x-3)-(x+1)(x+2)(x+3)}{(x+1)(x+2)(x+3)} >0\]
like you have to multiply (x-1)(x-2)(x-3) and multiply (x+1)(x+2)(x+3)
combine like terms on top
there might be a little trick to making a short cut
like if you already know the top isn't going to have any 0's
but I don't see how to do that without actually multiplying it all out and combing like terms
anyways the bottom is easy to see where we have vertical asymptotes
since (x+1)(x+2)(x+3)=0 is easy to solve

- freckles

I would draw a number line and test intervals around those vertical asymptotes and any 0's if they do exist

- mathmath333

there will come a cubic polynomial in numerator , how will i factor that

- freckles

you get a cubic poly in the numerator ?

- freckles

I think the x^3 will cancel in the top

- mathmath333

oh i see its not cubic

- freckles

\[(x-1)(x-2)(x-3) \\ =(x^2-3x+2)(x-3) \\ =x^3-3x^2+2x-3x^2-9x-6 \\ =x^3-6x^2-7x-6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\]
might check my arithmetic

- mathmath333

ok its easy nnow

- freckles

darn it

- freckles

I didn't multiply -3x(-3) correctly

- ikram002p

|dw:1436971504692:dw|

- freckles

\[(x-1)(x-2)(x-3) \\ =(x^2-3x+2)(x-3) \\ =x^3-3x^2+2x-3x^2+9x-6 \\ =x^3-6x^2+11x-6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\]

- ikram002p

|dw:1436971702003:dw|

- ikram002p

if you are not struggling with my graph then my solution would be
x>0 and \(x <\frac{ -1 }{ \sqrt[3]{2} }\)

- freckles

yep yep
on that one I liked multiplying both sides by x^2
and drawing a number line

- ikram002p

by x not x^2*

- freckles

\[2x^4+x>0 \\ x(2x^3+1)>0 \\ x(2x^3+1)=0 \text{ when } x=0 \text{ or } x=\frac{-1}{\sqrt[3]{2}} \approx -.79 \\ \]
|dw:1436972007081:dw|

- freckles

no by x^2

- freckles

you can also choose to multiply both sides by x but then you have cases

- freckles

we see that the function is only positive on (-inf,-1/cubeoot(2)) union (0,infty)
just as you got

- ikram002p

oh im referring to the origin didnt see other stuff :)

- freckles

I didn't actually solve it
I just suggested multiplying both sides by x^2 and that was all
I like to avoid those cases sometimes

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