A mathematical challenge for anyone who wants one?

Question

4y^2 +12xy+3x^2-2x+5=0. Find 2 algebraic functions that are determined by this equation. (ie, Use the quadratic formula to solve for y in terms of x).

Answer 1

y = 1/8(-12x +/- Sqrt(144x^2 - 16(3x^2- 2x + 5)))

Answer 2

4y^2 + 12xy + 3x^2 - 2x + 5 = 0

Complete the square.

4(y^2 + 3xy) + 3x^2 - 2x + 5 = 0
4(y^2 + 3xy + (9/4)x^2) + 3x^2 - 2x + 5 - (9/4)x^2 = 0
4(y + (3/2)x)^2 + 3x^2 - 2x + 5 - (9/4)x^2 = 0

4(y + (3/2)x)^2 + 3x^2 - (9/4)x^2 - 2x + 5 = 0

Group together the x^2 terms.

4(y + (3/2)x)^2 + x^2(3 - 9/4) - 2x + 5 = 0
4(y + (3/2)x)^2 + x^2(3/4) - 2x + 5 = 0

Move all terms with x in them to the right hand side.

4(y + (3/2)x)^2 = 2x - (3/4)x^2 - 5

Divide both sides by 4,

(y + (3/2)x)^2 = (1/2)x - (3/16)x^2 - (5/4)

Take the square root of both sides, yielding the two equations:

y + (3/2)x = (1/2)x - (3/16)x^2 - (5/4)
y + (3/2)x = - [(1/2)x - (3/16)x^2 - (5/4)]

Bringinng the (3/2)x to the other side in BOTH equations:

y = (1/2)x - (3/16)x^2 - (5/4) - (3/2)x
y = - [(1/2)x - (3/16)x^2 - (5/4)] - (3/2)x

After much simplification, your final two functions should be

y = (-3/16)x^2 - x - (5/4)
y = (3/16)x^2 - 2x + (5/4)