how can i prove the quadratic equation??

Question

homework

Answer 1

ax² + bx + c = 0

x² + bx/a + c/a = 0 dividing through by a

(x + b/(2a))² - b² / (4a²) + c/a = 0 completing the square

(x + b/(2a))² = b² / (4a²) - c/a

................... = b² / (4a²) - 4ac / (4a²)

................... = (b² - 4ac) / (4a²)

x + b/(2a) = ± √ [(b² - 4ac) / (4a²)]

................ = ± √(b² - 4ac) / 2a

x = - b/(2a) ± √(b² - 4ac) / 2a

x = [- b ± √(b² - 4ac)] / 2a

Answer 2

By completing the square of a x ^2 + b x in
a x ^2 + b x = c

Answer 3

Calculus or...

x^2 - 9
(x +3)(x -3)
x=-3 and 3

use the quadratic formula and u get the same.

Answer 4

ax² + bx + c = 0
x² + (b/a).x + c/a = 0
x² + (b/a).x + b²/4a² = b²/4a² - c/a
(x + b/2a)² = (b² - 4ac) / 4a²
x + b/2a = ±√(b² - 4ac) / 2a
x = - b/2a ±√ (b² - 4ac) /2a
x = [- b ± √(b² - 4ac)] / 2a
Example
2x² + 7x + 6 = 0
x = [ - 7 ± √(7² - 48) ] / 4
x = [ - 7 ± 1 ] / 4
x = - 6 / 4 , x = - 8 / 4
x = - 3/2 , x = - 2
By factorising:-
(2x + 3).(x + 2) = 0
x = - 3/2 , x = - 2 (as obtained by formula)

Hope this helps.

Answer 5

ax^2 + bx + c = 0
x^2 + bx/a + c/a = 0
x^2 + bx/a = -c/a
complete the square:
x^2 + bx/a + b^2/a^2 = b^2/4a^2 - c/a
(x + b/2a)^2 = b^2/4a^2 - 4ac/4a^2
x + b/2a = +/-(the square root of b^2-4ac)/2a
x = (-b+/-the square root of b^2-4ac)/2a

Answer 6

if a & b are the roots then: a + b = -m a * b = m^2 + n^2 a^4 + (ab)^2 + b^4 = a^4 + 2(ab)^2 + b^4 - (ab)^2 = (a^2 + b^2)^2 - (ab)^2 = (a^2 + b^2 - ab)(a^2 + b^2 + ab) = (a^2 + 2ab + b^2 - 3ab)(a^2 + 2ab + b^2 - ab) = [(a + b)^2 - 3ab] [(a + b)^2 - ab] = [(-m)^2 - 3(m^2 + n^2)] [(-m)^2 -(m^2 + n^2)] = (m^2 - 3m^2 - 3n^2)(m^2 - m^2 - n^2) = -n^2(-2m^2 - 3n^2) = n^2(2m^2 + 3n^2)

Answer 7

go to google, type in "proof quadratic equation" and up pops a bunch of sites.

http://www.math-help.info/Proof_of_Quadratic_Formula.html

Answer 8

well, if you have a squared term that's a good hint...

ax^2 + bx + c = D

a not equal 0

Good luck.