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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} 2x^2+\dfrac{1}{x}\gt0\hspace{.33em}\\~\\ \end{align}}\)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0multiply both side od inequality by x subtract 1 from both sides divide both sides by 2 take cube root

freckles
 one year ago
Best ResponseYou've already chosen the best response.2@MrNood if we multiply both sides by x. We have two cases. x is positive and x is negative.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.2if you want to avoid that can you try multiplying x^2 on both sides instead

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how to solve this \(x^3+\dfrac12>0\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2subtracting 1/2 on both sides then taking cube root of both sides

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u mean this ? \(\large \color{black}{\begin{align} x>\left(\dfrac12\right)^{1/3}\hspace{.33em}\\~\\ \end{align}}\) x can only have real values

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and since ()^(1/3) is an odd function you can actually bring the negative outside if you prefer

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[x> \frac{1}{2^\frac{1}{3}}\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but wolfram gives it as complex number http://www.wolframalpha.com/input/?i=%5Cleft%28%5Cdfrac12%5Cright%29%5E%7B1%2F3%7D%3D

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yes there are complex cube roots of 1/2 but you want the real cube root of 1/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=real+cube+root+of+%281%2F2%29

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0can this be done in case of square root too ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2what do you mean exactly ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0this one http://www.wolframalpha.com/input/?i=real+square+root+of+%281%2F2%29

freckles
 one year ago
Best ResponseYou've already chosen the best response.2short 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
 one year ago
Best ResponseYou've already chosen the best response.2though 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 yaxis so sqrt(1) is not going to have a real answer but cuberoot(x) function looks like this: dw:1436970632089:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the cube function actually does exist to the left of the yaxis

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u mean even roots dont have real roots

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I think I said square roots

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yes i did that is what I meant

freckles
 one year ago
Best ResponseYou've already chosen the best response.2square root of negative numbers aren't real

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i mean if \(\Large a^{1/b}\) if b is even then there is no real root ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=4th+roots+of+%281%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
 one year ago
Best ResponseYou've already chosen the best response.0all even roots are roots of the square root  so if the squre root is not real neither is any even root

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=6th+roots+of+%281%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
 one year ago
Best ResponseYou've already chosen the best response.2the example I showed though was of the square root of a negative number since you asked about sqrt(1/2)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok next qustion \(\large \color{black}{\begin{align} \dfrac{(x1)(x2)(x3)}{(x+1)(x+2)(x+3)}>1\hspace{.33em}\\~\\ \end{align}}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I would probably end up subtracting 1 on both sides and combining the fractions but that seems really long

freckles
 one year ago
Best ResponseYou've already chosen the best response.2only because there is so much multiplication we have to do in the numerator

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{(x1)(x2)(x3)(x+1)(x+2)(x+3)}{(x+1)(x+2)(x+3)} >0\] like you have to multiply (x1)(x2)(x3) 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
 one year ago
Best ResponseYou've already chosen the best response.2I would draw a number line and test intervals around those vertical asymptotes and any 0's if they do exist

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0there will come a cubic polynomial in numerator , how will i factor that

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you get a cubic poly in the numerator ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I think the x^3 will cancel in the top

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0oh i see its not cubic

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(x1)(x2)(x3) \\ =(x^23x+2)(x3) \\ =x^33x^2+2x3x^29x6 \\ =x^36x^27x6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\] might check my arithmetic

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I didn't multiply 3x(3) correctly

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436971504692:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(x1)(x2)(x3) \\ =(x^23x+2)(x3) \\ =x^33x^2+2x3x^2+9x6 \\ =x^36x^2+11x6 \\ (x+1)(x+2)(x+3) \\ =(x^2+3x+2)(x+3) \\ =x^3+3x^2+2x+3x^2+9x+6 \\ =x^3+6x^2+11x+6\]

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436971702003:dw

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1if you are not struggling with my graph then my solution would be x>0 and \(x <\frac{ 1 }{ \sqrt[3]{2} }\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep yep on that one I liked multiplying both sides by x^2 and drawing a number line

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[2x^4+x>0 \\ x(2x^3+1)>0 \\ x(2x^3+1)=0 \text{ when } x=0 \text{ or } x=\frac{1}{\sqrt[3]{2}} \approx .79 \\ \] dw:1436972007081:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you can also choose to multiply both sides by x but then you have cases

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we see that the function is only positive on (inf,1/cubeoot(2)) union (0,infty) just as you got

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1oh im referring to the origin didnt see other stuff :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I 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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