What are the solutions of the system of equations: #x^2+y-5=0#, #x+y^2-5=0# ?
1 Answer
Explanation:
Given:
#{ (x^2+y-5=0), (x+y^2-5=0) :}#
From the first equation, we can deduce that:
#y = 5-x^2#
Then substituting for
#0 = x+(5-x^2)^2-5#
#color(white)(0) = x^4-10x^2+x+20#
#color(white)(0) = (x^2-ax+b)(x^2+ax+c)#
#color(white)(0) = x^4+(b+c-a^2)x^2+a(b-c)x+bc#
Equating coefficients, we find:
#{ (b+c = a^2-10), (b-c=1/a), (bc=20) :}#
By inspection, this has solution:
#a = 1# ,#b = -4# ,#c = -5#
So:
#x^4-10x^2+x+20 = (x^2-x-4)(x^2+x-5)#
Using the quadratic formula, this has roots:
#x = 1/2+-sqrt(17)/2#
#x = -1/2+-sqrt(21)/2#
When
#y = 5-x^2#
#color(white)(y) = 5-(1/2+sqrt(17)/2)^2#
#color(white)(y) = 5-(1/4+sqrt(17)/2+17/4)#
#color(white)(y) = 1/2-sqrt(17)/2#
By symmetry we can deduce points of intersection:
#(1/2+sqrt(17)/2, 1/2-sqrt(17)/2)#
#(1/2-sqrt(17)/2, 1/2+sqrt(17)/2)#
When
#y = 5-x^2#
#color(white)(y) = 5-(-1/2+sqrt(21)/2)^2#
#color(white)(y) = 5-(1/2-sqrt(21)/2+21/4)#
#color(white)(y) = -1/2+sqrt(21)/2#
By symmetry we can deduce points of intersection:
#(-1/2+sqrt(21)/2, -1/2+sqrt(21)/2)#
#(-1/2-sqrt(21)/2, -1/2-sqrt(21)/2)#
graph{(x^2+y-5)(x+y^2-5) = 0 [-10, 10, -4.24, 5.76]}
