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formula that provides the solutions to a quadratic equation

A quadratic function with roots x = 1 and x = 4.

In

elementary algebra

, the quadratic formula is a formula that provides the solution(s) to a

. There are other ways of solving a quadratic equation instead of using the quadratic formula, such as

factoring

(direct factoring, grouping,

AC method

),

completing the square

,

graphing

and others.

[1]

Given a general quadratic equation of the form

ax2+bx+c=0{displaystyle ax^{2}+bx+c=0}

with x representing an unknown, a, b and c representing

constants

with a ≠ 0, the quadratic formula is:

x=−b2−4ac2a  {displaystyle x={frac {-bpm {sqrt {b^{2}-4ac}}}{2a}} }

where the

plus-minus symbol “±”

indicates that the quadratic equation has two solutions.

[2]

Written separately, they become:

x1=−b+b2−4ac2aandx2=−b−b2−4ac2a{displaystyle x_{1}={frac {-b+{sqrt {b^{2}-4ac}}}{2a}}quad {text{and}}quad x_{2}={frac {-b-{sqrt {b^{2}-4ac}}}{2a}}}

Each of these two solutions is also called a

root (or zero)

of the quadratic equation. Geometrically, these roots represent the x-values at which any

parabola

, explicitly given as y = ax2 + bx + c, crosses the x-axis.

[3]

As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola,

[4]

and the number of

real

zeros the quadratic equation contains.

[5]

## Equivalent formulations

The quadratic formula may also be written as

x=−b2a±b2−4ac4a2  ,{displaystyle x={frac {-b}{2a}}pm {sqrt {frac {b^{2}-4ac}{4a^{2}}}} ,}

which may be simplified to

x=−b2a±(b2a)2−ca  .{displaystyle x=-{frac {b}{2a}}pm {sqrt {left({frac {b}{2a}}right)^{2}-{frac {c}{a}}}} .}

The expression inside the square root is called the

discriminant

.
This version of the formula makes it easy to find the roots when using a calculator.
The version above is also convenient when complex roots are involved, in which case the expression outside the square root is the real part and the square root expression is the imaginary part:

x=− b2a±i|( b2a)2−ca|  .{displaystyle x=-{frac { b}{2a}}pm i{sqrt {{Big |}left({frac { b}{2a}}right)^{2}-{frac {c}{a}}{Big |}}} .}

Bạn đang xem: Quadratic formula-Wikipedia

### Muller’s method

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:

x=−2cb±b2−4ac=2c−b∓b2−4ac   .{displaystyle x={frac {-2c}{bpm {sqrt {b^{2}-4ac}}}}={frac {2c}{-bmp {sqrt {b^{2}-4ac }}}} .}

### Formulations based on alternative parametrizations

The standard parametrization of the quadratic equation is

ax2+bx+c=0  .{displaystyle ax^{2}+bx+c=0 .}

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

ax2−2b1x+c=0{displaystyle ax^{2}-2b_{1}x+c=0}, where b1=−b/2{displaystyle b_{1}=-b/2},

[6]

or

ax2+2b2x+c=0{displaystyle ax^{2}+2b_{2}x+c=0}, where b2=b/2{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.

## Derivations of the formula

Many different methods to derive the quadratic formula are available in the literature. The standard one is a simple application of the

completing the square

technique.

[8]

[9]

[10]

[11]

Alternative methods are sometimes simpler than completing the square, and may offer interesting insight into other areas of mathematics.

### By using the ‘completing the square’ technique

#### Standard method

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

x2+bax+ca=0  .{displaystyle x^{2}+{frac {b}{a}}x+{frac {c}{a}}=0 .}

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

x2+bax=−ca  .{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:

x2+bax+(b2a)2=−ca+(b2a)2  ,{displaystyle x^{2}+{frac {b}{a}}x+left({frac {b}{2a}}right)^{2}=-{frac {c}{a}}+left({frac {b}{2a}}right)^{2} ,}

which produces:

(x+b2a)2=−ca+b24a2  .{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:

(x+b2a)2=b2−4ac4a2  .{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:

x+b2a=±b2−4ac 2a  .{displaystyle x+{frac {b}{2a}}=pm {frac {sqrt {b^{2}-4ac }}{2a}} .}

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

x=−b2−4ac 2a  .{displaystyle x={frac {-bpm {sqrt {b^{2}-4ac }}}{2a}} .}

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

#### Method 2

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 a{displaystyle a} to make the polynomial

monic

.

2. Rearrange.
3. Add (b2a)2{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 x{displaystyle x}.

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

[12]

1. Multiply each side by 4a{displaystyle 4a},
2. Rearrange.
3. Add b2{displaystyle b^{2}} to both sides to complete the square.
4. Take the square root of both sides.
5. Isolate x{displaystyle x}.

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

ax2+bx+c=04a2x2+4abx+4ac=04a2x2+4abx=−4ac4a2x2+4abx+b2=b2−4ac(2ax+b)2=b2−4ac2ax+b=±b2−4ac2ax=−b2−4acx=−b2−4ac2a  .{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&=-bpm {sqrt {b^{2}-4ac}}\x&={frac {-bpm {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]

#### Method 3

Similar to Method 1, divide each side by a{displaystyle a} to make left-hand-side polynomial

monic

(i.e., the coefficient of x2{displaystyle x^{2}} becomes 1).

x2+bax+ca=0  .{displaystyle x^{2}+{frac {b}{a}}x+{frac {c}{a}}=0 .}

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

x2+Bx+C=0{displaystyle x^{2}+Bx+C=0}

where B=ba{displaystyle textstyle B={frac {b}{a}}} and C=ca{displaystyle textstyle C={frac {c}{a}}}.

Complete the square by adding B24{displaystyle textstyle {frac {B^{2}}{4}}} to the first two terms and subtracting it from the third term:

(x+B2)2+(C−B24)=0{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

:

(x+B2)2−(B24−C)2=0{displaystyle left(x+{frac {B}{2}}right)^{2}-left({sqrt {{frac {B^{2}}{4}}-C}}right)^{2}=0}

and factor it:

(x+B2+B24−C)(x+B2−B24−C)=0{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

x+B2+B24−C=0{displaystyle x+{frac {B}{2}}+{sqrt {{frac {B^{2}}{4}}-C}}=0}

or

x+B2−B24−C=0{displaystyle x+{frac {B}{2}}-{sqrt {{frac {B^{2}}{4}}-C}}=0}

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

x=−B2−B24−C{displaystyle x=-{frac {B}{2}}-{sqrt {{frac {B^{2}}{4}}-C}}} or x=−B2+B24−C{displaystyle x=-{frac {B}{2}}+{sqrt {{frac {B^{2}}{4}}-C}}}

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

citation needed

]

### By substitution

Another technique is solution by

substitution

.

[14]

In this technique, we substitute x=y+m{displaystyle x=y+m} into the quadratic to get:

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

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

ay2+y(2am+b)+(am2+bm+c)=0  .{displaystyle ay^{2}+y(2am+b)+left(am^{2}+bm+cright)=0 .}

We have not yet imposed a second condition on y{displaystyle y} and m{displaystyle m}, so we now choose m{displaystyle m} so that the middle term vanishes. That is, 2am+b=0{displaystyle 2am+b=0} or m=−b2a{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 a{displaystyle a} gives:

y2=−(am2+bm+c)a  .{displaystyle y^{2}={frac {-left(am^{2}+bm+cright)}{a}} .}

Substituting for m{displaystyle m} gives:

y2=−(b24a+−b22a+c)a=b2−4ac4a2  .{displaystyle y^{2}={frac {-left({frac {b^{2}}{4a}}+{frac {-b^{2}}{2a}}+cright)}{a}}={frac {b^{2}-4ac}{4a^{2}}} .}

Therefore,

y=±b2−4ac2a{displaystyle y=pm {frac {sqrt {b^{2}-4ac}}{2a}}}

By re-expressing y{displaystyle y} in terms of x{displaystyle x} using the formula x=y+m=y−b2a{displaystyle textstyle x=y+m=y-{frac {b}{2a}}} , the usual quadratic formula can then be obtained:

x=−b2−4ac2a  .{displaystyle x={frac {-bpm {sqrt {b^{2}-4ac}}}{2a}} .}

### By using algebraic identities

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:

(r1−r2)2=(r1+r2)2−4r1r2  .{displaystyle (r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}-4r_{1}r_{2} .}

Taking the square root on both sides, we get:

r1−r2=±(r1+r2)2−4r1r2  .{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,

x2+bax+ca=(x−r1)(x−r2)=x2−(r1+r2)x+r1r2  .{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:

r1−r2=±(−ba)2−4ca=±b2a2−4aca2=±b2−4aca  .{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,

r1=(r1+r2)+(r1−r2)2=−ba±b2−4aca2=−b2−4ac2a  .{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 {-bpm {sqrt {b^{2}-4ac}}}{2a}} .}

Since r2 = −r1b/a, if we take

r1=−b+b2−4ac2a{displaystyle r_{1}={frac {-b+{sqrt {b^{2}-4ac}}}{2a}}}

then we obtain

r2=−b−b2−4ac2a  ;{displaystyle r_{2}={frac {-b-{sqrt {b^{2}-4ac}}}{2a}} ;}

and if we instead take

r1=−b−b2−4ac2a{displaystyle r_{1}={frac {-b-{sqrt {b^{2}-4ac}}}{2a}}}

then we calculate that

r2=−b+b2−4ac2a  .{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:

x=−b2−4ac2a  .{displaystyle x={frac {-bpm {sqrt {b^{2}-4ac}}}{2a}} .}

### By Lagrange resolvents

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

x2+px+q  ,{displaystyle x^{2}+px+q ,}

assume that it factors as

x2+px+q=(x−α)(x−β)  ,{displaystyle x^{2}+px+q=(x-alpha )(x-beta ) ,}

Expanding yields

x2+px+q=x2−)x+αβ  ,{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 Sn. 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:

r1=αr2=αβ  .{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:

α=12(r1+r2)β=12(r1−r2)  .{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 = αβ is not symmetric, since switching α and β yields r2 = βα (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 = (αβ)2 is symmetric in the roots, and thus expressible in terms of p and q. Using the equation

β)2=(α)2−β  {displaystyle (alpha -beta )^{2}=(alpha +beta )^{2}-4alpha beta }

yields

r22=p2−4q  {displaystyle r_{2}^{2}=p^{2}-4q }

and thus

r2=±p2−4q  {displaystyle r_{2}=pm {sqrt {p^{2}-4q}} }

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

r1=−pr2=p2−4q{displaystyle {begin{aligned}r_{1}&=-p\r_{2}&={sqrt {p^{2}-4q}}end{aligned}}}

and thus

α=12(−p+p2−4q)β=12(−p−p2−4q)  .{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

12(−p2−4q){displaystyle textstyle {frac {1}{2}}left(-ppm {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.

## Historical development

The earliest methods for solving quadratic equations were geometric. Babylonian cuneiform tablets contain problems reducible to solving quadratic equations.

[18]

The Egyptian

Berlin Papyrus

, dating back to the

Middle Kingdom

(2050 BC to 1650 BC), contains the solution to a two-term quadratic equation.

[19]

The Greek mathematician

Euclid

(circa 300 BC) used geometric methods to solve quadratic equations in Book 2 of his

Elements

, an influential mathematical treatise.

[20]

Rules for quadratic equations appear in the Chinese

The Nine Chapters on the Mathematical Art

circa 200 BC.

[21]

[22]

In his work

Arithmetica

, the Greek mathematician

Diophantus

(circa 250 AD) solved quadratic equations with a method more recognizably algebraic than the geometric algebra of Euclid.

[20]

His solution gives only one root, even when both roots are positive.

[23]

The Indian mathematician

Brahmagupta

(597–668 AD) explicitly described the quadratic formula in his treatise

Brāhmasphuṭasiddhānta

published in 628 AD,

[24]

but written in words instead of symbols.

[25]

His solution of the quadratic equation ax2 + bx = c was as follows: “To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value.”

[26]

This is equivalent to:

x=4ac+b2−b2a  .{displaystyle x={frac {{sqrt {4ac+b^{2}}}-b}{2a}} .}

The 9th-century Persian mathematician

Muḥammad ibn Mūsā al-Khwārizmī

solved quadratic equations algebraically.

[27]

The quadratic formula covering all cases was first obtained by

Simon Stevin

in 1594.

[28]

In 1637

René Descartes

published

La Géométrie

containing special cases of the quadratic formula in the form we know today.

[29]

## Significant uses

### Geometric significance

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:

y=p(x)=a2x2+a1x+a0  ,{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,

x=−b2−4ac 2a=−b2a±b2−4ac 2a{displaystyle x={frac {-bpm {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.

### Dimensional analysis

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.

• Fundamental theorem of algebra

• Vieta’s formulas

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,

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,

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