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[parent] derivation of quadratic formula (Proof)

Suppose $ A,B,C$ are real numbers, with $ A\neq 0$, and suppose

$\displaystyle Ax^2+Bx+C=0.$

Since $ A$ is nonzero, we can divide by $ A$ and obtain the equation

$\displaystyle x^2 + bx+c = 0, $
where $ b=\frac{B}{A}$ and $ c=\frac{C}{A}$. This equation can be written as
$\displaystyle x^2 + bx + \frac{b^2}{4} -\frac{b^2}{4} +c =0,$
so completing the square, i.e., applying the identity $ (p+q)^2=p^2+2pq + q^2$, yields
$\displaystyle \left(x+\frac{b}{2}\right)^2 = \frac{b^2}{4} - c.$
Then, taking the square root of both sides, and solving for $ x$, we obtain the solution formula
$\displaystyle x$ $\displaystyle =$ $\displaystyle -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}-c}$  
  $\displaystyle =$ $\displaystyle \frac{B}{2A} \pm \sqrt{\frac{B^2}{4A^2}-\frac{C}{A}}$  
  $\displaystyle =$ $\displaystyle \frac{-B\pm\sqrt{B^2-4AC}}{2A},$  

and the derivation is completed.

A slightly less intuitive but more aesthetically pleasing approach to this derivation can be achieved by multiplying both sides of the equation

$\displaystyle ax^2+bx+c=0$    

by $ 4a$, resulting in the equation
$\displaystyle 4a^2x^2+4abx+b^2=b^2-4ac,$    

in which the left-hand side can be expressed as $ (2ax+b)^2$. From here, the proof is identical.



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See Also: quadratic formula, quadratic equation in $\mathbb{C}$


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Cross-references: proof, derivation, solution, sides, square root, identity, completing the square, equation, divide, real numbers
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This is version 8 of derivation of quadratic formula, born on 2001-11-07, modified 2006-07-22.
Object id is 700, canonical name is DerivationOfQuadraticFormula.
Accessed 19758 times total.

Classification:
AMS MSC12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )

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