# Need help Proving quadriatic equation

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

Staff Emeritus
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).