quadratic fundamentals?

Question

'everyone' knows that( -b^2+,-sqrt4ac etc ) thing,but I am having a ding dang of a time trying to work out how it is actually derived, and getting into a whole mess of algebraic trouble

Answer 1

it's done by completing the square.

ax^2 +bx + c =0 Divide by a
x^2 +b/a x + c/a = 0 Now remove c/a
x^2 + b/ax = -c/a We now can complete the square if we add
(b/2a)^2 to both sides
x^2 +b/2ax +(b/2a)^2 = b^2 /4a^2 -c/a
Factor the left side and combine the fractions on the right by making the denominators both 4a^2

(X + b/2a)^2 = (b^2 - 4ac)/4a^2
Take the square root of both sides
X + b/2a = plus or minus (sqrt(b^2 -4ac)/2a
just subtract -b/2a from both sides and you're there.

Answer 2

Given the quadratic equation:

ax² + bx + c = 0

where a ≠ 0

We can assume that a ≠ 0 without loss of generality because if it were equal to zero we wouldn't have a quadratic equation. Let's proceed with the derivation.

ax² + bx + c = 0

Divide by a

x² + (b/a)x + c/a = 0
x² + (b/a)x = -c/a

Complete the square.

x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
[x + b/(2a)]² = -c/a + (b/2a)² = -c/a + b²/(4a²)

[x + b/(2a)]² = (-4ca + b²) / (4a²)
[x + b/(2a)]² = (b² - 4ac) / (2a)²

Take the square root.

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

x = {-b ± √[(b² - 4ac)} / (2a)

Answer 3

ax^2+bx+c = 0
Multiply each term by a, if a≠0, and move ac to the right side,
(ax)^2+(abx) = -ac
Complete the square,
(ax)^2+(abx) + b^2/4 = b^2/4 - ac
(ax+b/2)^2 = (b^2-4ac)/4
Take the square root of both sides,
ax + b/2 = ±sqrt(b^2-4ac)/2
Solve for x,
x = [-b±sqrt(b^2-4ac)]/(2a)