## mathmath333 one year ago Solve

1. mathmath333

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

2. anonymous

x>0 is a solution.

3. MrNood

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

4. freckles

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

5. 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

6. freckles

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

7. mathmath333

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

8. freckles

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

9. 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

10. freckles

that is right

11. freckles

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

12. freckles

$x>- \frac{1}{2^\frac{1}{3}}$

13. mathmath333

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

14. freckles

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

15. freckles
16. mathmath333

can this be done in case of square root too ?

17. freckles

what do you mean exactly ?

18. mathmath333
19. 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

20. 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|

21. freckles

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

22. mathmath333

u mean even roots dont have real roots

23. freckles

I think I said square roots

24. freckles

yes i did that is what I meant

25. freckles

square root of negative numbers aren't real

26. mathmath333

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

27. 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

28. MrNood

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

29. 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

30. freckles

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

31. 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}}

32. freckles

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

33. freckles

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

34. 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

35. freckles

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

36. mathmath333

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

37. freckles

you get a cubic poly in the numerator ?

38. freckles

I think the x^3 will cancel in the top

39. mathmath333

oh i see its not cubic

40. 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

41. mathmath333

ok its easy nnow

42. freckles

darn it

43. freckles

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

44. ikram002p

|dw:1436971504692:dw|

45. freckles

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

46. ikram002p

|dw:1436971702003:dw|

47. ikram002p

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

48. freckles

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

49. ikram002p

by x not x^2*

50. 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|

51. freckles

no by x^2

52. freckles

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

53. freckles

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

54. ikram002p

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

55. 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