Why Does the Root Sum Calculation for a Polynomial's Derivative Confuse Many?

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The question is :"The polynomial 3x5-250x3+735x is interesting because it has the maximum possible number of relative extrema and points of inflection at integer lattice points for a quintic polynomial. What is the sum of the x-coordinates of these points?"

I think the answer is 0, but it's wrong.

My solution is :"The first derivative is 15x4-750x2+735, whose roots are -7,-1,1 and 7. The second derivative is 60x3-1500x,whose roots are 0,-5,5. Then, the sum is 0."

The HMMT solution is "The first derivative is 15x4-750x2+735, whose roots sum to 750/15=50. The second derivative is 60x3-1500x,whose roots sum to 1500/60=25, for a grand total of 75."

I really can't understand how it get the sum 50 and 25...Please help me.

Thanks...
 
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The only thing I can say is that they appear to be talking about the sum of x2 where x is a zero of the polynomial:
15x^4-750x^2+735= 15(x^2- 1)(x^2- 49)
so x2= 1 and 49 which add to 50 and
60x3-1500x= 60 x(x^2- 25)
so x2= 0 and 25 which add to 25.
 
HallsofIvy said:
The only thing I can say is that they appear to be talking about the sum of x2 where x is a zero of the polynomial:
15x^4-750x^2+735= 15(x^2- 1)(x^2- 49)
so x2= 1 and 49 which add to 50 and
60x3-1500x= 60 x(x^2- 25)
so x2= 0 and 25 which add to 25.

i think so...
but since we have +7 and -7 and so on...why isn't it 150?
 
thank you all the way...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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