Question

Solve

Answer 1

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

Answer 2

x>0 is a solution.

Answer 3

multiply both side od inequality by x

subtract 1 from both sides

divide both sides by 2

take cube root

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

how to solve this

\(x^3+\dfrac12>0\)

Answer 8

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

Answer 9

u mean this ?


\(\large \color{black}{\begin{align} x>\left(-\dfrac12\right)^{1/3}\hspace{.33em}\\~\\
\end{align}}\)

x can only have real values

Answer 10

that is right

Answer 11

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

Answer 12

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

Answer 13

but wolfram gives it as complex number

http://www.wolframalpha.com/input/?i=%5Cleft%28-%5Cdfrac12%5Cright%29%5E%7B1%2F3%7D%3D

Answer 14

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

Answer 15

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

Answer 16

can this be done in case of square root too ?

Answer 17

what do you mean exactly ?

Answer 18

this one

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

Answer 19

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

Answer 20

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:

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: