prove the quadratic formula without using completing the square?

Question

basically i need to go from ax^2+bx+c=0 to the quadratic formula without using completing the square. any suggestions?

Answer 1

The traditional proof involves completing the square. But you don't want us to do that, so let's try the direct approach: let x = (-b ± √(b² - 4ac))/(2a). Then we have that

ax² + bx + c

a((-b ± √(b² - 4ac))/(2a))² + b((-b ± √(b² - 4ac))/(2a)) + c

a((b² ∓ 2b√(b²-4ac) + b² - 4ac)/(4a²)) + b((-b ± √(b² - 4ac))/(2a)) + c

(2b² ∓ 2b√(b²-4ac) - 4ac)/(4a) + (-b² ± b√(b² - 4ac))/(2a) + c

(2b² ∓ 2b√(b²-4ac) - 4ac)/(4a) + (-2b² ± 2b√(b² - 4ac))/(4a) + c

(-4ac)/(4a) + c

-c + c

0

So (-b ± √(b² - 4ac))/(2a) is a solution of the equation ax² + bx + c = 0.

Answer 2

Prove Quadratic Formula

Answer 3

Michael Brozinsky:
The substitution z=x+b/(2a) transforms the standard quadratic equation ax^2+bx+c=0 into a quadratic equation in z missing a linear term. The roots for z are therefore opposites and the quadratic formula for x follows at once

Answer 4

it may provide help to ingredient the quadratic equation by using including yet another term which makes the equation have 3 words. it fairly is yet in any different case of factoring an quadratic equation with out having each and all of the words. It in basic terms works for quadratic equations,

Answer 5

PeterT still completed the square... when he went from lines 4 to 5 to 6.

He didnt do it in the traditional method... the simple method... that we all learn to do... but he still completed the square.

He artificially constructed a polynomial (by adding arbitrary terms to both sides) that resulted in an expression that could be factored into two identical factors

Answer 6

ax^2 + bx + c = 0
4a(ax^2 + bx + c) = 0
4a^2x^2 + 4abx + 4ac = 0
4a^2x^2 + 4abx = -4ac
4a^2x^2 + 4abx + b^2 = b^2 - 4ac
(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

QED