How do you find the exact solutions to the system #3x=8y^2# and #8y^2-2x^2=16#?
1 Answer
The possible solutions are:
#(x, y) = (1/4(3+sqrt(119)i), +-sqrt(3/32(3+sqrt(119)i)))#
#(x, y) = (1/4(3-sqrt(119)i), +-sqrt(3/32(3-sqrt(119)i)))#
Explanation:
Given:
#{ (3x=8y^2), (8y^2-2x^2=16) :}#
Substitute
#3x-2x^2=16#
Add
#0 = 2x^2-3x+16#
Multiply by
#0 = 16x^2-24x+128#
#color(white)(0) = (4x-3)^2+119#
#color(white)(0) = (4x-3)^2-(sqrt(119)i)^2#
#color(white)(0) = ((4x-3)-sqrt(119)i)((4x-3)+sqrt(119)i)#
#color(white)(0) = (4x-3-sqrt(119)i)(4x-3+sqrt(119)i)#
Hence:
#x = 1/4(3+-sqrt(119)i)#
Then:
#8y^2 = 3x = 3*1/4(3+-sqrt(119)i) = 3/4(3+-sqrt(119)i)#
Divide both ends by
#y^2 = 3/32(3+-sqrt(119)i)#
So the possible solutions are:
#(x, y) = (1/4(3+sqrt(119)i), +-sqrt(3/32(3+sqrt(119)i)))#
#(x, y) = (1/4(3-sqrt(119)i), +-sqrt(3/32(3-sqrt(119)i)))#
It is possible to express
Using https://socratic.org/s/aw38evei, the principal square root of
#(sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i#
I will let you plug in