1. Jun 19, 2004

princesscharming26

I need help proving the quadriatic equation... this is all i got up to:

ax(squared)+bx+(b/2)quantity squared= -c+(b/2)quantity squared

sorry.. i kind of dont know how to use the other codes!

2. Jun 19, 2004

AKG

Work your way backwards, then reverse the steps so you know the way forwards.

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Multiply both sides by 2a

$$2ax = -b \pm \sqrt{b^2 - 4ac}$$

$$2ax + b = \pm \sqrt{b^2 - 4ac}$$

Square both sides

$$4a^2x^2 + 4abx + b^2 = b^2 - 4ac$$

Subtract $b^2$ from both sides

$$4a^2x^2 + 4abx = - 4ac$$

$$4a^2x^2 + 4abx + 4ac = 0$$

Divide both sides by 4a

$$ax^2 + bx + c = 0 \ \dots \ (a \neq 0)$$

So, to go forwards, do the opposite of those actions in reverse order:

Multiply both sides by 4a
Subtract 4ac from both sides
Add $b^2$ to both sides
Square root both sides
Subtract b from both sides
Divide both sides by 2a

3. Jun 19, 2004

matt grime

Or better is to complete the square on the original eqaution.

4. Jun 20, 2004

StonedPanda

Exactly what I was going to say. Just complete the square with variables.

5. Jun 20, 2004

HallsofIvy

Actually, the original post was trying to complete the square. Unfortunately, she was doing it wrong:

After writing ax2+ bx= -c, divide both sides by a: x2+ (b/a)x= -c/a.

NOW complete the square: the coefficient of x is (b/a) so we square half of that and add (b/2a)2 to both sides:
x2+ (b/a)x+ (b/2a)2= (b/2a)2- c/a

princesscharming26, do you understand WHY you add that square?

It's because x2+ (b/a)x+ (b/2a)2= (x+ b/2a)2.

Now you have (x+ b/2a)2= b2/4a- c/a= (b2- 4ac)/(2a).