How do you solve the system #x^2+y^2=16# and #x^2-5y=5#?
1 Answer
Jul 29, 2016
#(+-sqrt(-15/2+(5sqrt(69))/2), -5/2+sqrt(69)/2)#
#(+-(sqrt(15/2+(5sqrt(69))/2))i, -5/2-sqrt(69)/2)#
Explanation:
Subtract the second equation from the first and rearrange to get:
#y^2+5y-11 = 0#
Use the quadratic formula to find:
#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
So we find
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)#