DERIVING QUADRATIC FORMULA FROM QUADRATIC EQUATION
Quadratic equation:
ax^2 + bx + c = 0
Transpose c to the right side
ax^2 + bx = -c
Divide both sides by a to eliminate a
x^2 + (b/a)x = -c/a
By completing the square:
Divide the coefficient of x by 2
Thus, (b/a) / 2 = b/2a
(x + b/2a)^2 = -c/a + (b/2a)^2
(x + b/2a)^2 = -c/a + (b^2)/(4a^2)
LCD of -c/a and (b^2)/(4a^2) is 4a^2
Thus, divide 4a^2 by each denominator then multiply to the numerator
(4a^2) / a = 4a
(4a)(-c) = -4ac
(4a^2) / (4a^2) = 1
(1)(b^2) = b^2
Thus,
(x + b/2a)^2 = (-4ac + b^2) / (4a^2)
Get the square root of both sides:
√(x + b/2a)^2 = +/- √(-4ac + b^2) / √(4a^2)
x + b/2a = +/- √(-4ac + b^2) / 2a
x = -b/2a +/- √(-4ac + b^2)/2a
Simplifying further, we will arrive at:
x = [-b +/- √(b^2 - 4ac)] / 2a
And that's how the quadratic formula was derived.
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-shinnichi
