How do you derive the quadratic formula from the pythagoream theorum?

Question

I don't get it.

Answer 1

You don't. It is derived from the "completion of the perfect square" method of solving for roots.

Answer 2

You don't. The quadratic formula is derived from the quadratic equation
ax^2 + bx + c = 0
as follows:
x^2 + (b/a)x + c/a = 0
x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2 + c/a = 0
(x + b/2a)^2 - b^2/4a^2 + 4ac/4a^2 = 0
(x + b/2a)^2 - (b^2 - 4ac)/4a^2 = 0
(x + b/2a - (√(b^2 - 4ac))/2a)(x + b/2a + (√(b^2 - 4ac))/2a) = 0
(x + (b - √(b^2 - 4ac))/2a)(x + (b + √(b^2 - 4ac))/2a) = 0
x = - (b - √(b^2 - 4ac))/2a, - (b + √(b^2 - 4ac))/2a
x = - (b ± √(b^2 - 4ac))/2a
or
x = (- b ± √(b^2 - 4ac))/2a

Answer 3

quadratic formula cannot be derived from this as in quadratic equations u have only 2 variables but pythagoras theorem have 3 variables............

Answer 4

(-b + square root of b^2 - 4ac)/2a and (-b - square root of b^2 - 4ac)/2a

plug in the values and get the answers