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

x>0 is a solution.

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

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

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

that is right

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

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

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

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

can this be done in case of square root too ?

what do you mean exactly ?

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

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

u mean even roots dont have real roots

I think I said square roots

yes i did
that is what I meant

square root of negative numbers aren't real

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

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

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

you get a cubic poly in the numerator ?

I think the x^3 will cancel in the top

oh i see its not cubic

ok its easy nnow

darn it

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

|dw:1436971504692:dw|

|dw:1436971702003:dw|

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

by x not x^2*

no by x^2

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

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

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