Quick question about a small problem

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Homework Statement



In my work I am stuck at this equation:
840+1.5v^{2.4}-3.6v^{1.4}=0

I am looking for a quick, easy way to solve this.
Any suggestions? I am kinda tired right now and nothing comes to my mind except maybe Newtons method?
 
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Okay, so this equation comes from the derivative of

s(x)=\frac{600v}{1.4+0.0025*v^{2.4}}

But Wolfram Alpha tells me the derivative is wrong? Oo
What did I do wrong?

s'(x)=\frac{600(1.4+0.0025v^{2.4})-600v(0.006v^{1.4})}{(1.4+0.0025*v^{2.4})^{2}}

s'(x)=\frac{840+1.5v^{2.4}-3.6v^{1.4}}{(1.4+0.0025*v^{2.4})^{2}}
 
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Elpinetos said:
Okay, so this equation comes from the derivative of

s(x)=\frac{600v}{1.4+0.0025*v^{2.4}}

But Wolfram Alpha tells me the derivative is wrong? Oo
What did I do wrong?

s'(x)=\frac{600(1.4+0.0025v^{2.4})-600v(0.006v^{1.4})}{(1.4+0.0025*v^{2.4})^{2}}

s'(x)=\frac{840+1.5v^{2.4}-3.6v^{1.4}}{(1.4+0.0025*v^{2.4})^{2}}

##v \times v^{1.4} = v^{2.4}##, not ##v^{1.4}##.
 
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Elpinetos said:

Homework Statement



In my work I am stuck at this equation:
840+1.5v^{2.4}-3.6v^{1.4}=0

I am looking for a quick, easy way to solve this.
Any suggestions? I am kinda tired right now and nothing comes to my mind except maybe Newtons method?

A simple plot of ##840+1.5v^{2.4}-3.6v^{1.4}## shows that the equation has no positive real roots and hence has no real roots at all.
 
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Thanks, I overlooked the v
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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