How do you solve the system #x^2+y^2=16# and #x^2-5y=5#?

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Answer 1 · 2016-07-29T21:29:23

#(+-sqrt(-15/2+(5sqrt(69))/2), -5/2+sqrt(69)/2)#

#(+-(sqrt(15/2+(5sqrt(69))/2))i, -5/2-sqrt(69)/2)#

Subtract the second equation from the first and rearrange to get:

#y^2+5y-11 = 0#

#y = (-5+-sqrt(5^2-4(1)(-11)))/2#

#=(-5+-sqrt(25+44))/2#

#=(-5+-sqrt(69))/2#

Then from the second equation, we have:

#x^2=5+5y = 5+5((-5+-sqrt(69))/2) = -15/2+-(5sqrt(69))/2#

Hence:

#x = +-sqrt(-15/2+-(5sqrt(69))/2)#

Note that #sqrt(69) > sqrt(64) = 8#, hence #(5sqrt(69))/2 > 15/2#

So we find #2# Real values of #x# and #2# non-Real Complex values of #x#.

So the solutions are:

#(+-(sqrt(-15/2+(5sqrt(69))/2)), -5/2+sqrt(69)/2)#

#(+-(sqrt(15/2+(5sqrt(69))/2))i, -5/2-sqrt(69)/2)#