Square root in the denominator

A lesser known quadratic formula, first mentioned by Giulio Fagnano,^[9] describes the same roots via an equation with the square root in the denominator (assuming ⁠${\displaystyle c\neq 0}$⁠):

$${\displaystyle x={\frac {2c}{-b\mp {\sqrt {b^{2}-4ac}}}}.}$$

Here the minus–plus symbol "⁠${\displaystyle \mp }$⁠" indicates that the two roots of the quadratic equation, in the same order as the standard quadratic formula, are

$${\displaystyle x_{1}={\frac {2c}{-b-{\sqrt {b^{2}-4ac}}}},\qquad x_{2}={\frac {2c}{-b+{\sqrt {b^{2}-4ac}}}}.}$$

This variant has been jokingly called the "citardauq" formula ("quadratic" spelled backwards).^[10]

When ⁠${\displaystyle -b}$⁠ has the opposite sign as either ⁠${\displaystyle \textstyle +{\sqrt {b^{2}-4ac}}}$⁠ or ⁠${\displaystyle \textstyle -{\sqrt {b^{2}-4ac}}}$⁠, subtraction can cause catastrophic cancellation, resulting in poor accuracy in numerical calculations; choosing between the version of the quadratic formula with the square root in the numerator or denominator depending on the sign of ⁠${\displaystyle b}$⁠ can avoid this problem. See § Numerical calculation below.

This version of the quadratic formula is used in Muller's method for finding the roots of general functions. It can be derived from the standard formula from the identity ⁠${\displaystyle x_{1}x_{2}=c/a}$⁠, one of Vieta's formulas. Alternately, it can be derived by dividing each side of the equation ⁠${\displaystyle \textstyle ax^{2}+bx+c=0}$⁠ by ⁠${\displaystyle \textstyle x^{2}}$⁠ to get ⁠${\displaystyle \textstyle cx^{-2}+bx^{-1}+a=0}$⁠, applying the standard formula to find the two roots ⁠${\displaystyle \textstyle x^{-1}\!}$⁠, and then taking the reciprocal to find the roots ⁠${\displaystyle x}$⁠ of the original equation.

Completing the square by Śrīdhara's method

Instead of dividing by ⁠${\displaystyle a}$⁠ to isolate ⁠${\displaystyle \textstyle x^{2}\!}$⁠, it can be slightly simpler to multiply by ⁠${\displaystyle 4a}$⁠ instead to produce ⁠${\displaystyle \textstyle (2ax)^{2}\!}$⁠, which allows us to complete the square without need for fractions. Then the steps of the derivation are:^[11]

1. Multiply each side by ⁠${\displaystyle 4a}$⁠.
2. Add ⁠${\displaystyle \textstyle b^{2}-4ac}$⁠ to both sides to complete the square.
3. Take the square root of both sides.
4. Isolate ⁠${\displaystyle x}$⁠.

Applying this method to a generic quadratic equation with symbolic coefficients yields the quadratic formula:

$${\displaystyle {\begin{aligned}ax^{2}+bx+c&=0\\[3mu]4a^{2}x^{2}+4abx+4ac&=0\\[3mu]4a^{2}x^{2}+4abx+b^{2}&=b^{2}-4ac\\[3mu](2ax+b)^{2}&=b^{2}-4ac\\[3mu]2ax+b&=\pm {\sqrt {b^{2}-4ac}}\\[5mu]x&={\dfrac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.{\vphantom {\bigg )}}\end{aligned}}}$$

This method for completing the square is ancient and was known to the 8th–9th century Indian mathematician Śrīdhara.^[12] Compared with the modern standard method for completing the square, this alternate method avoids fractions until the last step and hence does not require a rearrangement after step 3 to obtain a common denominator in the right side.^[11]

By substitution

Another derivation uses a change of variables to eliminate the linear term. Then the equation takes the form ⁠${\displaystyle \textstyle u^{2}=s}$⁠ in terms of a new variable ⁠${\displaystyle u}$⁠ and some constant expression ⁠${\displaystyle s}$⁠, whose roots are then ⁠${\displaystyle u=\pm {\sqrt {s}}}$⁠.

By substituting ⁠${\displaystyle x=u-{\tfrac {b}{2a}}}$⁠ into ⁠${\displaystyle \textstyle ax^{2}+bx+c=0}$⁠, expanding the products and combining like terms, and then solving for ⁠${\displaystyle \textstyle u^{2}\!}$⁠, we have:

$${\displaystyle {\begin{aligned}a\left(u-{\frac {b}{2a}}\right)^{2}+b\left(u-{\frac {b}{2a}}\right)+c&=0\\[5mu]a\left(u^{2}-{\frac {b}{a}}u+{\frac {b^{2}}{4a^{2}}}\right)+b\left(u-{\frac {b}{2a}}\right)+c&=0\\[5mu]au^{2}-bu+{\frac {b^{2}}{4a}}+bu-{\frac {b^{2}}{2a}}+c&=0\\[5mu]au^{2}+{\frac {4ac-b^{2}}{4a}}&=0\\[5mu]u^{2}&={\frac {b^{2}-4ac}{4a^{2}}}.\end{aligned}}}$$

Finally, after taking a square root of both sides and substituting the resulting expression for ⁠${\displaystyle u}$⁠ back into ⁠${\displaystyle x=u-{\tfrac {b}{2a}},}$⁠ the familiar quadratic formula emerges:

$${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$$

By using algebraic identities

The following method was used by many historical mathematicians:^[13]

Let the roots of the quadratic equation ⁠${\displaystyle \textstyle ax^{2}+bx+c=0}$⁠ be ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠. The derivation starts from an identity for the square of a difference (valid for any two complex numbers), of which we can take the square root on both sides:

$${\displaystyle {\begin{aligned}(\alpha -\beta )^{2}&=(\alpha +\beta )^{2}-4\alpha \beta \\[3mu]\alpha -\beta &=\pm {\sqrt {(\alpha +\beta )^{2}-4\alpha \beta }}.\end{aligned}}}$$

Since the coefficient ⁠${\displaystyle a\neq 0}$⁠, we can divide the quadratic equation by ⁠${\displaystyle a}$⁠ to obtain a monic polynomial with the same roots. Namely,

$${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=(x-\alpha )(x-\beta )=x^{2}-(\alpha +\beta )x+\alpha \beta .}$$

This implies that the sum ⁠${\displaystyle \alpha +\beta =-{\tfrac {b}{a}}}$⁠ and the product ⁠${\displaystyle \alpha \beta ={\tfrac {c}{a}}}$⁠. Thus the identity can be rewritten:

$${\displaystyle \alpha -\beta =\pm {\sqrt {\left(-{\frac {b}{a}}\right)^{2}-4{\frac {c}{a}}}}=\pm {\frac {\sqrt {b^{2}-4ac}}{a}}.}$$

Therefore,

$${\displaystyle {\begin{aligned}\alpha &={\tfrac {1}{2}}(\alpha +\beta )+{\tfrac {1}{2}}(\alpha -\beta )=-{\frac {b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac}}{2a}},\\[10mu]\beta &={\tfrac {1}{2}}(\alpha +\beta )-{\tfrac {1}{2}}(\alpha -\beta )=-{\frac {b}{2a}}\mp {\frac {\sqrt {b^{2}-4ac}}{2a}}.\end{aligned}}}$$

The two possibilities for each of ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ are the same two roots in opposite order, so we can combine them into the standard quadratic equation: $${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$$

By Lagrange resolvents

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents,^[14] which is an early part of Galois theory.^[15] This method can be generalized to give the roots of cubic polynomials and quartic polynomials, and leads to Galois theory, which allows one to understand the solution of algebraic equations of any degree in terms of the symmetry group of their roots, the Galois group.

This approach focuses on the roots themselves rather than algebraically rearranging the original equation. Given a monic quadratic polynomial ⁠${\displaystyle \textstyle x^{2}+px+q}$⁠ assume that ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ are the two roots. So the polynomial factors as

$${\displaystyle {\begin{aligned}x^{2}+px+q&=(x-\alpha )(x-\beta )\\[3mu]&=x^{2}-(\alpha +\beta )x+\alpha \beta \end{aligned}}}$$

which implies ⁠${\displaystyle p=-(\alpha +\beta )}$⁠ and ⁠${\displaystyle q=\alpha \beta }$⁠.

Since multiplication and addition are both commutative, exchanging the roots ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ will not change the coefficients ⁠${\displaystyle p}$⁠ and ⁠${\displaystyle q}$⁠: one can say that ⁠${\displaystyle p}$⁠ and ⁠${\displaystyle q}$⁠ are symmetric polynomials in ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠. Specifically, they are the elementary symmetric polynomials – any symmetric polynomial in ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ can be expressed in terms of ⁠${\displaystyle \alpha +\beta }$⁠ and ⁠${\displaystyle \alpha \beta }$⁠ instead.

The Galois theory approach to analyzing and solving polynomials is to ask whether, given coefficients of a polynomial each of which is a symmetric function in the roots, one can "break" the symmetry and thereby recover the roots. Using this approach, solving a polynomial of degree ⁠${\displaystyle n}$⁠ is related to the ways of rearranging ("permuting") ⁠${\displaystyle n}$⁠ terms, called the symmetric group on ⁠${\displaystyle n}$⁠ letters and denoted ⁠${\displaystyle S_{n}}$⁠. For the quadratic polynomial, the only ways to rearrange two roots are to either leave them be or to transpose them, so solving a quadratic polynomial is simple.

To find the roots ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠, consider their sum and difference:

$${\displaystyle r_{1}=\alpha +\beta ,\quad r_{2}=\alpha -\beta .}$$

These are called the Lagrange resolvents of the polynomial, from which the roots can be recovered as

$${\displaystyle \alpha ={\tfrac {1}{2}}(r_{1}+r_{2}),\quad \beta ={\tfrac {1}{2}}(r_{1}-r_{2}).}$$

Because ⁠${\displaystyle r_{1}=\alpha +\beta }$⁠ is a symmetric function in ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠, it can be expressed in terms of ⁠${\displaystyle p}$⁠ and ⁠${\displaystyle q,}$⁠ specifically ⁠${\displaystyle r_{1}=-p}$⁠ as described above. However, ⁠${\displaystyle r_{2}=\alpha -\beta }$⁠ is not symmetric, since exchanging ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ yields the additive inverse ⁠${\displaystyle -r_{2}=\beta -\alpha }$⁠. So ⁠${\displaystyle r_{2}}$⁠ cannot be expressed in terms of the symmetric polynomials. However, its square ⁠${\displaystyle \textstyle r_{2}^{2}=(\alpha -\beta )^{2}}$⁠ is symmetric in the roots, expressible in terms of ⁠${\displaystyle p}$⁠ and ⁠${\displaystyle q}$⁠. Specifically ⁠${\displaystyle \textstyle r_{2}^{2}=(\alpha -\beta )^{2}={}}$⁠⁠${\displaystyle \textstyle (\alpha +\beta )^{2}-4\alpha \beta ={}}$⁠⁠${\displaystyle \textstyle p^{2}-4q}$⁠, which implies ⁠${\displaystyle \textstyle r_{2}=\pm {\sqrt {p^{2}-4q}}}$⁠. Taking the positive root "breaks" the symmetry, resulting in

$${\displaystyle r_{1}=-p,\qquad r_{2}={\textstyle {\sqrt {p^{2}-4q}}}}$$

from which the roots ⁠${\displaystyle \alpha }$⁠ and ⁠${\displaystyle \beta }$⁠ are recovered as

$${\displaystyle x={\tfrac {1}{2}}(r_{1}\pm r_{2})={\tfrac {1}{2}}{\bigl (}{-p}\pm {\textstyle {\sqrt {p^{2}-4q}}}\,{\bigr )}}$$

which is the quadratic formula for a monic polynomial.

Substituting ⁠${\displaystyle p=b/a}$⁠, ⁠${\displaystyle q=c/a}$⁠ yields the usual expression for an arbitrary quadratic polynomial. The resolvents can be recognized as

$${\displaystyle {\tfrac {1}{2}}r_{1}=-{\tfrac {1}{2}}p=-{\frac {b}{2a}},\qquad r_{2}^{2}=p_{2}-4q={\frac {b^{2}-4ac}{a^{2}}},}$$

respectively the vertex and the discriminant of the monic polynomial.

A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating ⁠${\displaystyle r_{2}}$⁠ and ⁠${\displaystyle r_{3}}$⁠, which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved.^[14] The same method for a quintic equation yields a polynomial of degree 24, which does not simplify the problem, and, in fact, solutions to quintic equations in general cannot be expressed using only roots.