ax² + bx + c = 0 -> x = [-b ±sqr(b²-4ac)]/2a, demonstration?

Question

Can someone give me the demonstration that leads to the quadratic equation? (or where to find it)

Answer 1

basically you require a proof.

multiply by 4a to make 1st term a whole square and avoid a fraction
(2ax)^2 +4abx + 4ac =0
add b^2 -b^2 to make 1st part square
(2ax)^2+4abx +b^2 - b^2 +4ac =0
or (2ax+b)^2 = b^2-4ac
(2ax+b) = (+/-)sqrt(b^2-4ac)
=> 2ax = -b (+/-) sqrt(b^2-4ac)
=> x = (-b(+/-) sqrt(b^2-4ac))/2a

Answer 2

For all real numbers a, b, and c, a ≠ 0:

ax² + bx + c = 0

Let x = y + n

a(y + n)² + b(y + n) + c = 0

a(y² + 2ny + n²) + by + bn + c = 0

ay² + 2any + an² + by + bn + c = 0

ay² + (2an + b)y + an² + bn + c = 0

Let 2an + b = 0 or n = -b/(2a)

ay² + a(-b/2a)² + b(-b/2a) + c = 0

ay² + a(b²/4a²) - b²/(2a) + c = 0

ay² + b²/(4a) - b²/(2a) + c = 0

ay² + (b²/a)(1/4 - 1/2) + c = 0

ay² - b²/(4a) + 4ac/4a = 0

ay² - (1/4a)(b² - 4ac) = 0

y² - (1/4a²)(b² - 4ac) = 0

[y + (1/2a)√(b² - 4ac)] [y - (1/2a)√(b² - 4ac)]= 0

y = x - n = x - (-b/(2a)) = x + b/(2a)

[x + b/(2a) + (1/2a)√(b² - 4ac)] [x + b/(2a) - (1/2a)√(b² - 4ac)]= 0

Therefore:

x = [-b + √(b² - 4ac)]/(2a) or x = [-b - √(b² - 4ac)]/(2a)

Answer 3

ax^2+bx+c=0

ax^2+bx= -c

x^2 + b/a x = -c/a

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

( x + b/2a) ^2 = b^2-4ac/4a^2

x+ b/2a = + or - square root of b^2-4ac/2a

so x= {-b [+ or -] square root of b^2-4ac }/2a

Answer 4

http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html

Answer 5

I dont know