Solve bx + c = 0 for x by means of the quadratic formula.
Haven't looked at the others but these 3 occurred to me. Non of them are really calculus:
1) Multiply both sides of the equation by x.
Results in bx^2 + cx = 0 which can be solved by the quadratic formula. x = (-c +/- c)/2b
Solution x=0 can be discarded leaving x=-c/b
2) Square both sides of the equation.
Results in b^2x^2 + 2bcx + c^2 = 0 which can be solved by the quad. form. x = (-2bc +/- 0)/(2b^2)
Solution is a double root x=-c/b
3) Substitute y^2 for x.
Results in by^2 + c which can be solved by the quad. form.
y = (0 +/- sqrt(-4bc))/2b = +/- sqrt(-bc)/b = +/- i sqrt (bc)/b
y^2 = -b^2c^2/b^2 = -c/b
(I realize the first two can be generalized to multiplying both sides by x - h and discarding the extraneous root x = h)
Posted by Jer
on 2007-10-24 08:31:53