Derive the Quadratic equation from standard form (ax^2+bx+c=0) must show work?

ax^2 + bx + c = 0





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





remember that if (x+y)^2 = x^2 + 2xy + y^2





so, if you assume 'b/a x' as middle term for (x+y)^2,





we can write a ( x^2 + b/ax + (b/2a)^2 - (b/2a)^2 ) + c = 0.





be careful that x^2 + b/ax+ (b/2a)^2 mean (x+ b/2a)^2.





so, rewrite that a { (x+b/2a)^2 - (b/2a)^2 } + c = 0.





so, a( x+b/2a)^2 - b/4a +c = 0.





generally, we assume h= b/2a and k= c- b/4a





in this way, we can transform ax^2 + bx + c into a(x+h)^2+k.














again, ax^2+bx+c = 0,.





ax^2 + bx = -c





x^2 + b/ax = -c/a


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





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





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





so x = [-b +or- sqrt { (b^2-4ac) } ] / 2a.








this is the solution to all quadratic equation.








note b^2 is always %26gt; 4ac.





if b^2%26lt; 4ac, then we get complex numbers of roots.

Derive the Quadratic equation from standard form (ax^2+bx+c=0) must show work?
I guess you didn't look around very much as even Wikipedia gives the derivation.
Reply:http://en.wikipedia.org/wiki/Quadratic_e...

phlox