Solve #{(x^2+y^2= 2),(x^3+y^3 = 2):}# ?

Question

715 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-29T21:18:03

#a+b = 2#, or #a+b = 0.7322#

graphical solution:

[note: here, #a# and #b# have been substituted for #x# and #y#]
desmos.com/calculator

the (circular) graph in red shows #x^2+y^2 = 2#
the graph in blue shows #x^3+y^3 = 2#

the points of intersection are:
#(1,1)#
#(-0.5645, 1.2967)#
#(1.2967, -0.5645)#
rounded to #4# decimal places.

adding these values together gives either #2# or #0.7322#, for #a+b#.

Answer 2 · 2018-01-29T21:49:39

See below.

We have

#(a+b)^2 = a^2+b^2+2 a b = 2+2 a b#

#(a^3+b^3)/(a+b) = a^2-a b + b^2 = 2-a b# or

#2 = (a+b)(2-a b)#

Now calling #y = a+b# we have

#{(y^2= 2 + 2 ab),(2 = y/(2-ab)):}#

now equating #ab# in both equations we get

#y^3-6y+4=0# and solving we obtain

#y = a+b = {(2),(-(1+sqrt3)),(sqrt3-1):}#