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

1 Answer
George C.
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.

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