Thread: Make Y the subject

2(y)^2 - 3y = x + 2

You should recognize this as a Quadratic Equation and proceed. Completing the Square would be a nice, instructive method.

Less instructive: use the quadratic formula to solve the equation /

Less instructive yet(!): let k = x + 2; then: 2y^2 - 3y - k = 0

2(Y)^2 - 3Y - K = 0 (K = X+2)

Y^2 - (3/2)Y - K/2 = 0

(Y - 3/4)(Y - 3/4) - K/2 - 9/16 = 0

(Y - 3/4)^2 - (K/2 + 9/16) = 0

(Y - 3/4)^2 = (K/2 + 9/16)

Y - 3/4 = sqrt(K/2 + 9/16)

Y = 3/4 + sqrt(K/2 + 9/16)

Y = 3/4 + sqrt((X + 2)/2 + 9/16)

Y = 3/4 + sqrt((8X + 16)/16 + 9/16)

Y = 3/4 + sqrt((8X + 16 + 9)/16)

Y = sqrt((8X + 25)/16)

Y = (3 +/- sqrt(8X + 25))/4

Originally Posted by ajwright

2(Y)^2 - 3Y - K = 0 (K = X+2)

Y^2 - (3/2)Y - K/2 = 0

(Y - 3/4)(Y - 3/4) - K/2 - 9/16 = 0

(Y - 3/4)^2 - (K/2 + 9/16) = 0

(Y - 3/4)^2 = (K/2 + 9/16)

Y - 3/4 = +/- sqrt(K/2 + 9/16)

Y = 3/4 +/- sqrt(K/2 + 9/16)

Y = 3/4 +/- sqrt((X + 2)/2 + 9/16)

Y = 3/4 +/- sqrt((8X + 16)/16 + 9/16)

Y = 3/4 +/- sqrt((8X + 16 + 9)/16)

Y = 3/4 +/- sqrt((8X + 25)/16)

Y = (3 +/- sqrt(8X + 25))/4

You need to try very hard make sure every step of your working is correct. See the corrections in red. The final answer is correct though, well done.

WHY do all that work? Never heard of the quadratic formula?

2y^2 - 3y - k = 0 , where k = x + 2

y = {-(-3) +/- SQRT[(-3)^2 - 4(2)(-k)]} / 4

y = [3 +/- SQRT(9 + 8k)] / 4
Substitute k = x + 2:
y = [3 +/- SQRT(9 + 8(x + 2))] / 4

y = [3 +/- SQRT(8x + 25)] / 4

Done!