Hidden symmetry

Algebra Level 4

Find the smallest root of the equation x 3 + 677 x 2 25 x 16925 = 0 x^{3}+677x^{2}-25x-16925=0 .


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The answer is -677.

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4 solutions

Spandan Senapati
Mar 18, 2017

We could use 16925 = 25 677 16925=25*677 Then the polynomial factorises to x 3 + 677 x 2 25 x 25 677 = ( x + 677 ) ( x + 5 ) ( x 5 ) x^3+677x^2-25x-25*677=(x+677)(x+5)(x-5) So smallest Root is 677 -677

Rohith M.Athreya
Feb 8, 2017

f ( x , y ) = ( 1 + x + y 2 ) ( 1 + y + x 2 ) = x 3 + x 2 ( 1 + y 2 ) + x ( 1 + y ) + y 3 + y 2 + y + 1 \displaystyle f(x,y)=(1+x+y^{2})(1+y+x^{2})=x^{3}+x^{2}(1+y^{2})+x(1+y)+y^{3}+y^{2}+y+1

simply set y = 3 i y=3i to get first expression and y = 26 y=-26 to get the second.

The roots are 1 y , 1 y , 1 y 2 \sqrt{-1-y},-\sqrt{-1-y},-1-y^{2}

Rohith buddy you have to change the terminology you have used

Anirudh Chandramouli - 4 years, 4 months ago

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yep,done :)

Rohith M.Athreya - 4 years, 4 months ago

It's written above 'hidden symmetry'. We can easily see (by using calculator) that 5 and -5 satisfy the eqn. Then by using sum of roots>>>>> 5+(-5)+c= -677 => c= -677

Michael Mendrin
Feb 8, 2017

Seems a lot easier solving the equation directly, instead of trying to 1) figure out what "...look for some homogeneity" means, and 2) actually finding that "homogeneity", whatever that means. That seems like a very difficult way to solve such an equation.

How does that method help find roots of an equation?

I felt the same, but @RohithMAthreya's solution also seems fine

Anirudh Chandramouli - 4 years, 4 months ago

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I'm just asking about how helpful it would be for other such polynomials.

Michael Mendrin - 4 years, 4 months ago

I dont know how to solve cubic equations in a way other than first guessing a root and then solving a quadratic(which according to me is not a great method)

i was thinking if some symmetry could be established,one could do away with the trial and error.

as for what is easier, i can't say with certainty as to which method wins as it is perception dependent

this method would help find a solution to a class of cubic equations (very narrow,nevertheless) i am looking to broaden the class with some more work and this was only a precursor.

a person is most definitely free to solve the equation by itself but might also try to look around for some symmetry(which may seem abstract) if he has the time and patience

Rohith M.Athreya - 4 years, 4 months ago

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and yes, homogeneity was a wrong term ( i was mistaken about what homogeneity was)

Rohith M.Athreya - 4 years, 4 months ago

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probably "homogeneity" is the wrong word to use here, but it's nevertheless an interesting concept here.

Michael Mendrin - 4 years, 4 months ago

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@Michael Mendrin I am glad you think so :)

Rohith M.Athreya - 4 years, 4 months ago

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