How do you find the exact solutions to the system #3x=8y^2# and #8y^2-2x^2=16#?

1 Answer
Jan 21, 2017

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# for #8y^2# in the second equation to get:

#3x-2x^2=16#

Add #2x^2-3x# to both sides to get:

#0 = 2x^2-3x+16#

Multiply by #8# to find:

#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 #8# to find:

#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 #sqrt(3/32(3-sqrt(119)i))# in #a+bi# form if you want, but it gets a little messier:

Using https://socratic.org/s/aw38evei, the principal square root of #a+bi# if it is in Q1 is:

#(sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i#

I will let you plug in #a=9/32# and #b=(3sqrt(119))/32# if you like.