derive the quadratic equation. start with a general form of a quadratic equation and complete the square?

Question

THANKS!!

Answer 1

General form:
Ax² + Bx + C = 0

Complete the square:
Ax² + Bx = -C
x² + (B/A)x = -(C/A)

Now, we know that x + 2k + k² = (x + k)².
So, let 2k = B/A, then add k² to both sides of our equation:
B/A = 2k
k = B/(2A)
k² = B²/(4A²)
x² + (B/A)x + B²/(4A²) = B²/(4A²) - C/A

Now factor it and solve for x:
[x + B/(2A)]² = B²/(4A²) - (4AC)/(4A²)
x + B/2A = ±sqrt[(B² - 4AC)/(4A²)]
x = -B/2A ± sqrt(B² - 4AC)/(2A)
x = [-B ± sqrt(B² - 4AC)]/(2A)
There you have it... the quadratic formula.

Answer 2

it truly is the quadratic formula. think you've the equation in the shape ax^2 + bx + c = 0, for constants a, b, and c. Then, to sparkling up for x, you take advantage of the quadratic formula. x = [-b +/- sqrt(b^2 - 4ac)]/(2a) The "+/-" shows that you'll doubtlessly get 2 solutions.

Answer 3

Given a, b, c:

ax² + bx + c
= (sqrta*x + k)² + constant
= ax² + 2k*sqrta*x + k² + constant

Setting equal coefficients for the 2nd term we have
b = 2k*sqrta
→ k = b/(2sqrta)

And so your quadratic becomes
= ax² + bx + k² + constant
= ax² + bx + b²/(4a) + constant

Thus your completed square will be
ax² + bx + c
= ax² + bx + c + b²/(4a) - b²/(4a)
= (ax² + bx + b²/(4a)) + (c - b²/(4a))
= [sqrta*x + b/(2sqrta)]² + (c - b²/(4a))

Answer 4

ax^2 +bx +c =0
ax^2+bx = - c
x^2 + b/a x = -c/a
x^2 +b/a x + b^2/4a^2 = -c/a +b^2/4a^2
(x+b/2a)^2 = (b^2-4ac)/4a^2
x+b/2a = +/- sqrt (b^2-4ac)/2a
x = -b/2a +/- sqrt(b^2-4ac)/2a
x = [-b +/- sqrt(b^2-4ac)]/2a