Oh my god!

Algebra Level 5

Find the sum of the roots of the following equation.

$x^{8061}+\left(\dfrac{1}{2}-x\right)^{8061}=0$

The answer is 2015.

2 solutions

Otto Bretscher
Jan 2, 2016

Make a substitution $x=y+\frac{1}{4}$ and write the equation as $(\frac{1}{4}+y)^{8061}+(\frac{1}{4}-y)^{8061}=0$ . The sum of the roots of this even polynomial of degree 8060 is 0, so that the sum of the roots of the original polynomial is $\frac{8060}{4}=\boxed{2015}$ .

I calculated the coefficient of $x^{8061}$ and $x^{8060}$ and $x^{8059}$ in the expansion of $(\frac {1}{2}-x)^{8061}$ and used Vieta to find the sum of the roots though I think your solution seems very elegant.

Anupam Nayak - 5 years, 5 months ago

Hello,

Maybe I don't understand the question very well because I'm French, but this equation doesn't have any real root... x^8061=-(1/2-x)^8061 <=> x=-1/2+x <=> 0=-1/2 <=> S=0 => sum(x,xES)=sum(x,xE0)=0... isn't it?

Elden Knight - 5 years, 5 months ago

True, the roots will all be non-real!

Otto Bretscher - 5 years, 5 months ago

Thanks! :) I have understood, so... :D

Elden Knight - 5 years, 5 months ago

But are we sure that the roots are simple roots?... because if not, if roots are roots of the derivative polynom, we don't have 8060 terms in the sum but less...................

Elden Knight - 5 years, 5 months ago

There are no multiple roots in this case... just take the derivative and solve the system $f(x)=f'(x)=0$ .

When people ask for the sum of the roots of a polynomial, they usually mean the sum of the complex roots, counted with their multiplicities... maybe this should have been stated explicitly.

Otto Bretscher - 5 years, 5 months ago

OK... I didn't know for the multiplicitie count... Thanks.

Elden Knight - 5 years, 5 months ago

@Elden Knight – Well, there are no multiple roots, so, it's not really an issue here...

Otto Bretscher - 5 years, 5 months ago

sir please elaborate

Deepansh Jindal - 4 years, 10 months ago