 How do I show that ax²+bx+c=0 can be written in the quadratic formula
But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead Original post by Bird1234
How do I show that ax²+bx+c=0 can be written in the quadratic formula
But don't get why they did that. I understand it involves completing the square but wouldn't that mean it should be (x+b/a)²- b²/a²+c/a=0 instead

Dividing through by a:

x^2 + b/a x + c/a = 0

Then when you complete the square for e.g. x^2+4x+5 you halve the b coefficient to give (x+2)^2 - 4 + 5

So in this case you halve b/a = b/2a to give (x+(b/2a))^2 - (b/2a)^2 + c/a Original post by Notnek
Dividing through by a:

x^2 + b/a x + c/a = 0

Then when you complete the square for e.g. x^2+4x+5 you halve the b coefficient to give (x+2)^2 - 4 + 5

So in this case you halve b/a = b/2a to give (x+(b/2a))^2 - (b/2a)^2 + c/a

Ohh OK thank you Alternatively, reverse engineer by expanding the result also works...