Find the number of solutions in real numbers to the system of equations below.
x4+2y3-x=-1/4+3√3
y4+2x3-y=-1/4-3√3
If we sum both
(x4+2x3-x)+(y4+2y3-y)=-1/2 equation (1)
We call them f(x)+f(y)=-1/2
If we get the roots of f(x)
f(x)=x(x+1)(x-0.6180)(x+1.6180)
There are four real roots, so we can have an idea of how is going f(x)
The derivative f'(x)=0 gives three points, a maximum for x=-0,5 and two minimum for x=0.366 and x= -1.366
The value of f(x) for this points are
x=0.366 f(x)=-0.25
x=-0.5 f(x)=0.312
x=-1.366 f(x)=-0.25
But then the minimum posible value for f(x)+f(y) is when both functions are -0.25. Then the sum is -1/2 which is the sum of equation (1).
From the enunciate is clear that x=y is not a solution, so the only possible solution is when
f(x)=-0.25 (x=-1.366) and f(y)=-0.25 (y=0.366)
Edited on March 18, 2019, 7:35 pm
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Posted by armando
on 2019-03-18 19:24:56 |
Solution
