Quadratic formula

A lesser known quadratic formula, which is used in Muller's method and which can be found from Vieta's formulas, provides the same roots via the equation:

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

The standard parametrization of the quadratic equation is

${\displaystyle ax^{2}+bx+c=0\ \ .}$

Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as

${\displaystyle ax^{2}-2b_{1}x+c=0}$, where ${\displaystyle b_{1}=-b/2}$,[6]

or

${\displaystyle ax^{2}+2b_{2}x+c=0}$, where ${\displaystyle b_{2}=b/2}$.[7]

These alternative parametrizations result in slightly different forms for the solution, but which are otherwise equivalent to the standard parametrization.

Divide the quadratic equation by ${\displaystyle a}$, which is allowed because ${\displaystyle a}$ is non-zero:

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0\ \ .}$

Subtract c/a from both sides of the equation, yielding:

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

The quadratic equation is now in a form to which the method of completing the square is applicable. In fact, by adding a constant to both sides of the equation such that the left hand side becomes a complete square, the quadratic equation becomes:

${\displaystyle x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+\left({\frac {b}{2a}}\right)^{2}\ \ ,}$

which produces:

${\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+{\frac {b^{2}}{4a^{2}}}\ \ .}$

Accordingly, after rearranging the terms on the right hand side to have a common denominator, we obtain:

${\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}\ \ .}$

The square has thus been completed. Taking the square root of both sides yields the following equation:

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

In which case, isolating the ${\displaystyle x}$ would give the quadratic formula:

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

There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of ${\displaystyle a}$.

The majority of algebra texts published over the last several decades teach completing the square using the sequence presented earlier:

1. Divide each side by ${\displaystyle a}$ to make the polynomial monic.
2. Rearrange.
3. Add ${\displaystyle \textstyle \left({\frac {b}{2a}}\right)^{2}}$ to both sides to complete the square.
4. Rearrange the terms in the right side to have a common denominator.
5. Take the square root of both sides.
6. Isolate ${\displaystyle x}$.

Completing the square can also be accomplished by a sometimes shorter and simpler sequence:[12]

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

In which case, the quadratic formula can also be derived as follows:

${\displaystyle {\begin{aligned}ax^{2}+bx+c&=0\\4a^{2}x^{2}+4abx+4ac&=0\\4a^{2}x^{2}+4abx&=-4ac\\4a^{2}x^{2}+4abx+b^{2}&=b^{2}-4ac\\(2ax+b)^{2}&=b^{2}-4ac\\2ax+b&=\pm {\sqrt {b^{2}-4ac}}\\2ax&=-b\pm {\sqrt {b^{2}-4ac}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ .\end{aligned}}}$

This derivation of the quadratic formula is ancient and was known in India at least as far back as 1025.[13] Compared with the derivation in standard usage, this alternate derivation avoids fractions and squared fractions until the last step and hence does not require a rearrangement after step 3 to obtain a common denominator in the right side.[12]

Similar to Method 1, divide each side by ${\displaystyle a}$ to make left-hand-side polynomial monic (i.e., the coefficient of ${\displaystyle x^{2}}$ becomes 1).

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0\ \ .}$

Write the equation in a more compact and easier-to-handle format:

${\displaystyle x^{2}+Bx+C=0}$

where ${\displaystyle \textstyle B={\frac {b}{a}}}$ and ${\displaystyle \textstyle C={\frac {c}{a}}}$.

Complete the square by adding ${\displaystyle \textstyle {\frac {B^{2}}{4}}}$ to the first two terms and subtracting it from the third term:

${\displaystyle \left(x+{\frac {B}{2}}\right)^{2}+\left(C-{\frac {B^{2}}{4}}\right)=0}$

Rearrange the left side into a difference of two squares:

${\displaystyle \left(x+{\frac {B}{2}}\right)^{2}-\left({\sqrt {{\frac {B^{2}}{4}}-C}}\right)^{2}=0}$

and factor it:

${\displaystyle \left(x+{\frac {B}{2}}+{\sqrt {{\frac {B^{2}}{4}}-C}}\right)\left(x+{\frac {B}{2}}-{\sqrt {{\frac {B^{2}}{4}}-C}}\right)=0}$

which implies that either

${\displaystyle x+{\frac {B}{2}}+{\sqrt {{\frac {B^{2}}{4}}-C}}=0}$

or

${\displaystyle x+{\frac {B}{2}}-{\sqrt {{\frac {B^{2}}{4}}-C}}=0}$

Each of these two equations is linear and can be solved for ${\displaystyle x}$, obtaining:

${\displaystyle x=-{\frac {B}{2}}-{\sqrt {{\frac {B^{2}}{4}}-C}}}$ or ${\displaystyle x=-{\frac {B}{2}}+{\sqrt {{\frac {B^{2}}{4}}-C}}}$

By re-expressing ${\displaystyle B}$ and ${\displaystyle C}$ back into ${\displaystyle \textstyle {\frac {b}{a}}}$ and ${\displaystyle \textstyle {\frac {c}{a}}}$, respectively, the quadratic formula can then be obtained.

Another technique is solution by substitution.[14] In this technique, we substitute ${\displaystyle x=y+m}$ into the quadratic to get:

${\displaystyle a(y+m)^{2}+b(y+m)+c=0\ \ .}$

Expanding the result and then collecting the powers of ${\displaystyle y}$ produces:

${\displaystyle ay^{2}+y(2am+b)+\left(am^{2}+bm+c\right)=0\ \ .}$

We have not yet imposed a second condition on ${\displaystyle y}$ and ${\displaystyle m}$, so we now choose ${\displaystyle m}$ so that the middle term vanishes. That is, ${\displaystyle 2am+b=0}$ or ${\displaystyle \textstyle m={\frac {-b}{2a}}}$. Subtracting the constant term from both sides of the equation (to move it to the right hand side) and then dividing by ${\displaystyle a}$ gives:

${\displaystyle y^{2}={\frac {-\left(am^{2}+bm+c\right)}{a}}\ \ .}$

Substituting for ${\displaystyle m}$ gives:

${\displaystyle y^{2}={\frac {-\left({\frac {b^{2}}{4a}}+{\frac {-b^{2}}{2a}}+c\right)}{a}}={\frac {b^{2}-4ac}{4a^{2}}}\ \ .}$

Therefore,

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

By re-expressing ${\displaystyle y}$ in terms of ${\displaystyle x}$ using the formula ${\displaystyle \textstyle x=y+m=y-{\frac {b}{2a}}}$ , the usual quadratic formula can then be obtained:

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

The following method was used by many historical mathematicians:[15]

Let the roots of the standard quadratic equation be r₁ and r₂. The derivation starts by recalling the identity:

${\displaystyle (r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}-4r_{1}r_{2}\ \ .}$

Taking the square root on both sides, we get:

${\displaystyle r_{1}-r_{2}=\pm {\sqrt {(r_{1}+r_{2})^{2}-4r_{1}r_{2}}}\ \ .}$

Since the coefficient a ≠ 0, we can divide the standard equation by a to obtain a quadratic polynomial having the same roots. Namely,

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=(x-r_{1})(x-r_{2})=x^{2}-(r_{1}+r_{2})x+r_{1}r_{2}\ \ .}$

From this we can see that the sum of the roots of the standard quadratic equation is given by −b/a, and the product of those roots is given by c/a. Hence the identity can be rewritten as:

${\displaystyle r_{1}-r_{2}=\pm {\sqrt {\left(-{\frac {b}{a}}\right)^{2}-4{\frac {c}{a}}}}=\pm {\sqrt {{\frac {b^{2}}{a^{2}}}-{\frac {4ac}{a^{2}}}}}=\pm {\frac {\sqrt {b^{2}-4ac}}{a}}\ \ .}$

Now,

${\displaystyle r_{1}={\frac {(r_{1}+r_{2})+(r_{1}-r_{2})}{2}}={\frac {-{\frac {b}{a}}\pm {\frac {\sqrt {b^{2}-4ac}}{a}}}{2}}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ .}$

Since r₂ = −r₁ − b/a, if we take

${\displaystyle r_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}}$

then we obtain

${\displaystyle r_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\ \$ ;}

and if we instead take

${\displaystyle r_{1}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}}$

then we calculate that

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

Combining these results by using the standard shorthand ±, we have that the solutions of the quadratic equation are given by:

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

An alternative way of deriving the quadratic formula is via the method of Lagrange resolvents,[16] which is an early part of Galois theory.[17] 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 more than on rearranging the original equation. Given a monic quadratic polynomial

${\displaystyle x^{2}+px+q\ \ ,}$

assume that it factors as

${\displaystyle x^{2}+px+q=(x-\alpha )(x-\beta )\ \ ,}$

Expanding yields

${\displaystyle x^{2}+px+q=x^{2}-(\alpha +\beta )x+\alpha \beta \ \ ,}$

where p = −(α + β) and q = αβ.

Since the order of multiplication does not matter, one can switch α and β and the values of p and q will not change: one can say that p and q are symmetric polynomials in α and β. In fact, they are the elementary symmetric polynomials – any symmetric polynomial in α and β can be expressed in terms of α + β and αβ The Galois theory approach to analyzing and solving polynomials is: given the coefficients of a polynomial, which are symmetric functions in the roots, can one "break the symmetry" and recover the roots? Thus solving a polynomial of degree n is related to the ways of rearranging ("permuting") n terms, which is called the symmetric group on n letters, and denoted S_n. For the quadratic polynomial, the only way to rearrange two terms is to swap them ("transpose" them), and thus solving a quadratic polynomial is simple.

To find the roots α and β, consider their sum and difference:

${\displaystyle {\begin{aligned}r_{1}&=\alpha +\beta \\r_{2}&=\alpha -\beta \ \ .\end{aligned}}}$

These are called the Lagrange resolvents of the polynomial; notice that one of these depends on the order of the roots, which is the key point. One can recover the roots from the resolvents by inverting the above equations:

${\displaystyle {\begin{aligned}\alpha &=\textstyle {\frac {1}{2}}\left(r_{1}+r_{2}\right)\\\beta &=\textstyle {\frac {1}{2}}\left(r_{1}-r_{2}\right)\ \ .\end{aligned}}}$

Thus, solving for the resolvents gives the original roots.

Now r₁ = α + β is a symmetric function in α and β, so it can be expressed in terms of p and q, and in fact r₁ = −p as noted above. But r₂ = α − β is not symmetric, since switching α and β yields −r₂ = β − α (formally, this is termed a group action of the symmetric group of the roots). Since r₂ is not symmetric, it cannot be expressed in terms of the coefficients p and q, as these are symmetric in the roots and thus so is any polynomial expression involving them. Changing the order of the roots only changes r₂ by a factor of −1, and thus the square r₂² = (α − β)² is symmetric in the roots, and thus expressible in terms of p and q. Using the equation

${\displaystyle (\alpha -\beta )^{2}=(\alpha +\beta )^{2}-4\alpha \beta \ \ }$

yields

${\displaystyle r_{2}^{2}=p^{2}-4q\ \ }$

and thus

${\displaystyle r_{2}=\pm {\sqrt {p^{2}-4q}}\ \ }$

If one takes the positive root, breaking symmetry, one obtains:

${\displaystyle {\begin{aligned}r_{1}&=-p\\r_{2}&={\sqrt {p^{2}-4q}}\end{aligned}}}$

and thus

${\displaystyle {\begin{aligned}\alpha &={\tfrac {1}{2}}\left(-p+{\sqrt {p^{2}-4q}}\right)\\\beta &={\tfrac {1}{2}}\left(-p-{\sqrt {p^{2}-4q}}\right)\ \ .\end{aligned}}}$

Thus the roots are

${\displaystyle \textstyle {\frac {1}{2}}\left(-p\pm {\sqrt {p^{2}-4q}}\right)}$

which is the quadratic formula. Substituting p = b/a, q = c/a yields the usual form for when a quadratic is not monic. The resolvents can be recognized as r₁/2 = −p/2 = −b/2a being the vertex, and r₂² = p² − 4q is the discriminant (of a monic polynomial).

A similar but more complicated method works for cubic equations, where one has three resolvents and a quadratic equation (the "resolving polynomial") relating r₂ and r₃, 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.[16] 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.

Graph of y = ax² + bx + c, where a and the discriminant b² − 4ac are positive, with

• Roots and y-intercept in red
• Vertex and axis of symmetry in blue
• Focus and directrix in pink

In terms of coordinate geometry, a parabola is a curve whose (x, y)-coordinates are described by a second-degree polynomial, i.e. any equation of the form:

${\displaystyle y=p(x)=a_{2}x^{2}+a_{1}x+a_{0}\ \ ,}$

where p represents the polynomial of degree 2 and a₀, a₁, and a₂ ≠ 0 are constant coefficients whose subscripts correspond to their respective term's degree. The geometrical interpretation of the quadratic formula is that it defines the points on the x-axis where the parabola will cross the axis. Additionally, if the quadratic formula was looked at as two terms,

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

the axis of symmetry appears as the line x = −b/2a. The other term, √b² − 4ac/2a, gives the distance the zeros are away from the axis of symmetry, where the plus sign represents the distance to the right, and the minus sign represents the distance to the left.

If this distance term were to decrease to zero, the value of the axis of symmetry would be the x value of the only zero, that is, there is only one possible solution to the quadratic equation. Algebraically, this means that √b² − 4ac = 0, or simply b² − 4ac = 0 (where the left-hand side is referred to as the discriminant). This is one of three cases, where the discriminant indicates how many zeros the parabola will have. If the discriminant is positive, the distance would be non-zero, and there will be two solutions. However, there is also the case where the discriminant is less than zero, and this indicates the distance will be imaginary – or some multiple of the complex unit i, where i = √−1 – and the parabola's zeros will be complex numbers. The complex roots will be complex conjugates, where the real part of the complex roots will be the value of the axis of symmetry. There will be no real values of x where the parabola crosses the x-axis.

If the constants a, b, and/or c are not unitless, then the units of x must be equal to the units of b/a, due to the requirement that ax² and bx agree on their units. Furthermore, by the same logic, the units of c must be equal to the units of b²/a, which can be verified without solving for x. This can be a powerful tool for verifying that a quadratic expression of physical quantities has been set up correctly, prior to solving this.