Can you put y=ax^2+bx+c in standard form by completing the square, I know this seems weird see datais?

the result of completing the square for the equation will be a shortcut in which you can just plug in the numbers and solve

Can you put y=ax^2+bx+c in standard form by completing the square, I know this seems weird see datais?
y = ax^2 + bx + c


y/a = x^2 + (b/a)x + (c/a)


y/a = x^2 + (b/a)x + (c/a) + (b/2a)^2 - (b/2a)^2


y/a = (x + b/2a)^2 + c/a - (b/2a)^2


y = a[(x + b/2a)^2 + c/a - (b/2a)^2]
Reply:The result of completing the square will be the "quadratic formula". There are many references for this. I included one for you.
Reply:When you complete the square you halve the co-efficeient of X, and take it's square outside the brackets, and add the constant on. I've only done it with numbers but I think it would be





a(X-1/2b)^2+b^2+c all over a





I'm not sure you're meant to sub numbers into it, but anyway to complete the square, you halve co-efficent of X, and put in brackets like so


(X-1/2X)^2+X^2+C


((you add the sqr and constant back on, the first X comes from the X^2)


The aX^2+bX+C can be used for other polynomial funtions like factor theorem, if that's what you're doing you'll need to say so...
Reply:ax^2 + bx + c = 0





divide by a





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





x^2 + 2(1)((b/2a) x + b^2/4a^2 - b^2/4a^2 + c/a = 0





(x + b/2a)^2 = b^2/4a^2 - c/a





(x + b/2a)^2 = (b^2 - 4a^2c/a)/4a^2





(x + b/2a)^2 = (b^2 - 4ac/)/4a^2





(x+b/2a) = +/- sqrt(b^2 - 4ac)/2a





x = -b/2a +/- sqrt(b^2-4ac)/2a





x = [ -b +/- sqrt(b^2 - 4ac)]/2a