+53 I know the quadratic formula, but I don't get why it works. Can someone post a proof of the quadratic formula.
x=−b±√b2−4ac2a
a*x^2 +b*x + c = 0.
Solve for x by completing the square:
divide the equation by 'a'
x^2 + b*x/a = -c/a
compare with (x+d)^2 = (x+d)*(x+d) = x^2 + 2d*x + d^2
so we need to find 'd' such that
2d = b/a or d = b/2a
and d^2 = b^2/4a^2
so, let's add 'd^2' to both sides of the equation
x^2 + d + d^2 = -c/a +d^2
or
(x+d)^2 = -c/a + d^2
take the square root of both sides and substitute in for 'd'
x+d = (+ or -) sqrt(-c/a + d^2)
x + b/4a = sqrt(-c/a + b^2/4a^2) = sqrt((b^2 - 4ac)/4a^2) = sqrt(b^2 - 4ac)/2a
x = -b/4a + sqrt(b^2 - 4ac)/2a = -b/4a + (+ or -)sqrt(b^2 - 4ac)/4a
x = [-b +/- sqrt(b^2 - 4ac)]
a*x^2 +b*x + c = 0.
Solve for x by completing the square:
divide the equation by 'a'
x^2 + b*x/a = -c/a
compare with (x+d)^2 = (x+d)*(x+d) = x^2 + 2d*x + d^2
so we need to find 'd' such that
2d = b/a or d = b/2a
and d^2 = b^2/4a^2
so, let's add 'd^2' to both sides of the equation
x^2 + d + d^2 = -c/a +d^2
or
(x+d)^2 = -c/a + d^2
take the square root of both sides and substitute in for 'd'
x+d = (+ or -) sqrt(-c/a + d^2)
x + b/4a = sqrt(-c/a + b^2/4a^2) = sqrt((b^2 - 4ac)/4a^2) = sqrt(b^2 - 4ac)/2a
x = -b/4a + sqrt(b^2 - 4ac)/2a = -b/4a + (+ or -)sqrt(b^2 - 4ac)/4a
x = [-b +/- sqrt(b^2 - 4ac)]
