Isolating a variable in a multivariable equation?

Question

I'm trying to derive a formula for my calculus class but the algebra has me stuck...I'm trying to solve the equation for x but can't get rid of the xsub1...Im dealing with points in the coordinate plane so I have xsub1 and ysub1 in the equation, here I'll just replace them with h and k, respectively...thanks so much in advance!

x^2 - x - xh = (4(Bk + C) + A)/(4A) - 1/4

Answer 1

x^2 - x - xh = (4(Bk + C) + A)/(4A) - 1/4

Let RHS = K
so x^2 - (1+h)x - K = 0

x = [ (1+h) +or- sqrt( (1+h)^2 - 4(-K) ) ] / 2
= [ (1+h) +or- sqrt( (1+h)^2 + 4K ) ] / 2

Answer 2

xsub1 and ysub1 are both constants, and xsub1 should be treated as a different letter from x. So replacing these with h and k is fine.

x^2 - (1 + h)x = D, where D = (4(Bk + C) + A)/(4A) - 1/4

Then x^2 - (1 + h)x - D = 0, so

x = (1 + h +/- sqrt((1 + h)^2 + 4D))/2

I hope this is enough.

Answer 3

write it as
x² - (1+h)x = [4Bk + 4C + A]/(4A) - 1/4
x² - (1+h)x = Bk/A + C/A + 1/4 - 1/4
x² - (1+h)x = Bk/A + C/A
and think about completing the square by adding the square of (1+h)/2
[x - (1+h)/2]² = Bk/A + C/A + (1+h)²/4
x - (1+h)/2 = ±√[Bk/A + C/A + (1+h)²/4]
x = (1+h)/2 ± √[Bk/A + C/A + (1+h)²/4]

Answer 4

x^2 - x - xh = (4(Bk + C) + A)/(4A) - 1/4
x^2 -x(1-h) -(4(Bk + C) + A)/(4A) + 1/4
x = [-b +/- sqrt(b^2-4ac)]/(2a) where:
a = 1
b = 1-h
c = -(4(Bk + C) + A)/(4A) + 1/4
x = [h-1 +/- sqrt((1-h)^2 - 4(1)(-4(Bk + C) + A)/(4A) + 1/4)]/2
x = [h-1 +/- sqrt(1-2h+h^2+16Bk+16C+A)/(4A) +1/4)]/2
You can simplify this a little bit, but it's not worth the effort.