How can I derive the quadratic formula?

Question

How can I derive the quadratic formula?

Answer 1

Consider the process of completing the square on a general quadratic equation:

ax² + bx + c = 0

First, turn it into a monic equation by dividing by a:

x² + (b/a)x + c/a = 0

For this to be a perfect square trinomial, the constant term must be (b/(2a))². So make it that by adding (b/(2a))² - c/a to both sides:

x² + (b/a)x + (b/(2a))² = (b/(2a))² - c/a

Factor the polynomial on the left, and collect the terms on the right:

(x + b/(2a))² = b²/(4a²) - c/a
(x + b/(2a))² = b²/(4a²) - 4ac/(4a²)
(x + b/(2a))² = (b² - 4ac)/(4a²)

Take the square roots of both sides:

x + b/(2a) = ±√(b²-4ac)/(2a)

Subtract b/(2a) from both sides:

x = (-b ± √(b² - 4ac))/(2a)

And we are done.

Answer 2

ax^2+bx+c=0 move c to the right side ax^2 + bx = -c divide through a, the coefficient of x^2 x^2 + (b/a)x = -c/a take half of b/a, the coefficient of x, and square it (b/a) / 2 = b/2a (b/2a)^2 = b^2/4a^2 then add it to both sides of the equation x^2 + (b/a)x + b^2/4a^2 = -c/a + b^2/4a^2 complete the square on the left side and simplify the right side (x + b/2a)^2 = (-4ac + b^2) /4a^2 take the square root of both sides x + b/2a = +/- [sqrt (b^2 -4ac)] / 2a add -b/2a to both sides of the equation x = -b/2a +/- [sqrt (b^2 -4ac)] / 2a simplify the left side x = [ - b +/- sqrt (b^2 - 4ac)] / 2a tala! now you have the formula:)

Answer 3

Start with ax^2 + bx + c = 0
Divide by a ("a" is not zero)

x^2 + (b/a)x + (c/a) = 0

Move the (c/a) to the right side
x^2 +(b/a)x = -(c/a)

Complete the square
x^2 + (b/a)x + b^2/(4a^2) = -(c/a) + b^2/(4a^2)
( x + b/(2a) )^2 = (b^2 - 4ac) / (4a^2)

Square root of each side
x + b/(2a) = (+/-) sqrt( b^2 - 4ac) / (2a)

Subtract b/(2a) from each side
x = (+/-) sqrt( b^2 - 4ac) / (2a) - b/(2a)

Collect terms
x = [ b (+/-) sqrt( b^2 - 4ac) ] / (2a)

Done!

Answer 4

You derive it by completing the square with the general formula of a quadratic equation.
ax²+bx+c = 0
x² + bx/a + c/a = 0
x² + bx/a + (b/(2a))² - (b/(2a))² + c/a = 0
(x + b/(2a))² = (b/(2a))² - c/a
(x + b/(2a))² = b²/(4a²) - (4ac)/(4a²)
(x + b/(2a))² = (b²-4ac)/(4a²)
x + b/(2a) = √(b²-4ac) / ±2a
x = (-b ± √(b²-4ac) ) / 2a

Answer 5

Set up your equation as
ax^2 + bx + c = 0.
Now, a (x^2 + b/a x +.................) + c - .................. = 0
complete the square of the expression in parenthesis, then subtract a * what you added from C. Work with it.