By The Fundamental Theorem of Algebra, an algebraic equation of the nth degree has n roots, all determined by solution of the equation. Since this equattion is constructed by applying the operations of arithmetic in some order, it follows that the solution can proceed by UNDOING of those operations by applying its INVERSE in the appropriate order. This can be illustrated by a familiar case, studied in high school algebra.

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DERIVATION OF THE QUADRATIC FORMULA

This derivation demonstrates that "algebra is arithmetic backward", as well as the usefulness of the bypass strategy.

• Our goal is a quadratic formula with "algenraic term", x, on the left-side of an equational form, and literal constants (surrogates for numbers) on the right-side.
• We start with the general quadratic equation:

  		ax2 + bx + c = 0 (1)

• We need to simplify it as a "monic" equation with leading coefficient of unity. Since the highest term of the equation, x², has a multiplier of a, we undo this by the inverse operation of division, dividing (1) through by a:

  		x2 + (b/a)x + c/a = 0 (2)

• To obtain that "goal" form, described above, we further simplify (2) by subtracting the constant term of literals, c/a, from both sides of (2):

  		x2 + (b/a)x = -(c/a) (3)

• We now need to reduce this quadratic to a linear equation by calling upon the inverse of the operation making it quadratic, namely, the inverse of exponentiation which is square-rooting.
• We know that square-rooting is easiest for a binomial form, such as u² + 2uv + v² = (u + v)², whose square root is u + v.
• Equation (3) does not have that form, but can be recast in that binomial form:

  		(x + k)2 = x2 + 2kx + k1 = 0 (4)

• The term 2kx of the expanded part of (4) corresponds to term (b/a)x in (3). So the term k in the closed binomial form of (4) corresponds to (b/2a), in terms of equation (3).
• So we consider the bypass of (3) by considering a closed binomial form:

  (x + b/2a)2 = x2 + (b/2a)x + (b/2a)x + b2/4a2 = x2 + (b/a)x + b2/4a2.(5)

• We notice that (5) differs from the left-side of (3) by having an adjoined term, b²/4a². We can maintain equality by adding such a term to the right-side of (3), obtaining:

  (x + b/2a)2 = b2/4a2 - c/a. (5)

• The left-side of (5) is "square-rootable", but the right-side is only partially so. We need the two right-side terms to have a least common denominator, namely, 4a².
• We again resort to "backwards arithmetic", multiplying both numerator and denominator of term -c/a, to become the equivalent term -4ac/4a², obtaining:

          (x + b/2a)2 = b2/4a2 - 4ac/a2 = (b2 - 4ac)/4a2. (6)

• Now, We can obtain a linear form by taking the square root of both sides of (6):

  		x + b/2a = ±((b2 - 4ac)/4a2) = ±((b2 - 4ac))/2a (7) 

• But we need a quadratic formula solution with only x on the left-side. Again via "backwards arithmetic", by subtracting the (added) term of b/2a from both sides, obtaining the standard form of the quadratic equation:

                  x + b/2a = b/2a ±((b2 - 4ac))/2a  = (b ±(b2 - 4ac))/2a (8) 

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  Note: This "trick" of (1) recasting in a form for square-rooting, and (2) performing the square root operation was emulated (in 1928) by British mathematical physicist, P. A. M. Dirac (x-y) -- in his relativistic theory of the electron -- to obtain a (linear) relativistic wave equation from the ("relativistic") Klein-Gordon wave equation (second order in space, but first-order in time) derived from Schrödinger's (nonrelativistic) wave equation.

  After putting the Klein=Gordon equation in suitable form (comparable to steps (1)-(6) above), Dirac used (without fully realizing it) a matrix operation equivalent to the multivector multiproduct (PL). In textbooks, the clumsiness of matrices requires about four confusing pages to explain this "sqaure-rooting" -- to achieve a wave equation that is (relativistically) first-order in both space and time, and containing spin -- whereas the "square-rooting" can be clearly written in four lines via the multiproduct.