Why does the following nonlinear system have 21 solutions? y + x y 2 </msup>

slijmigrd

slijmigrd

Answered question

2022-07-05

Why does the following nonlinear system have 21 solutions?
y + x y 2 x 3 + 2 x z 4 = 0 x y 3 3 x 2 y + 3 y z 4 = 0 5 2 y 2 z 3 2 x 2 z 3 z 7 2 = 0
there is one real and twenty complex solutions: ( 0 , 0 , 0 )

Answer & Explanation

Gornil2

Gornil2

Beginner2022-07-06Added 20 answers

Mark the solution ( 0 , 0 , 0 ) and remove z 3 from the last equation. The reduced systems for z = 0 has two cubic equations, so at most 3 3 = 9 solutions.Then z only occurs as z 4 . Multiply the first equation by x, the second by y and express all equations in a = x 2 , b = y 2 , c = x y and d = z 4 , adding the equation a b c 2 = 0. Then the system reads as
c + c 2 a 2 + 2 a d = 0 a b c 3 a c + 3 c d = 0 5 b 4 a d = 0 a b c 2 = 0
This system has the Bezout bound 8 for the number of solutions for ( a , b , c , d ), each of these will have at most 8 solutions for ( x , y , z ). This reduces the estimate to 64 + 9 + 1 = 74 solutions from the 5 5 7 = 175 or 5 5 4 + 3 3 = 109 for the Bezout bound of the (reduced) original system.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?