Need help Proving quadriatic equation

[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

:frown:

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

[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}

Add b to both sides

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

Add 4ac to both sides

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

[matt grime]

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

[StonedPanda]

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

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

[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).