Question

Coolest. Quadratic. Formula. Proof. Evar.

Answer 1

I discovered this on another Q/A sites. We have\[ax^2 + bx + c = 0\]Multiply both sides by \(4a\).\[4a^2x^2 + 2abx + 4ac = 0\]Factor.\[(2ax + b)^2 - b^2 = 4ax^2 + 2abx + 4ac\]We can write the above as\[(2ax + b)^2 - b^2 + 4ac = 0\]Or,\[(2ax + b)^2 = b^2 - 4ac\]Square root both sides.\[2ax + b = \pm\sqrt{b^2 - 4ac}\]Therefore,\[x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}\]

Answer 2

Brilliant ! parth minor typo there in x term... after myltiplying 4a both sides.. 2abx -> 4abx