In mathematics, a quadratic equation is a polynomial equation of the second degree. The general form is

$ax^2+bx+c=0,\backslash ,$

where

$a\backslash ne\; 0.\; \backslash ,$

The letters a, b, and c are called coefficients: the quadratic coefficient a is the coefficient of x², the linear coefficient b is the coefficient of x, and c is the constant coefficient, also called the free term.

Quadratic equations are called quadratic because quadratus is Latin for "square"; in the leading term the variable gets squared.

Quadratic formula

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A quadratic equation with real or complex coefficients has two complex roots (i.e., solutions for $x\; \backslash$) denoted here as $x\_+\; \backslash$ and $x\_-\; \backslash$, although the two roots may be equal. These roots $x\_\backslash pm$ can be computed using the quadratic formula:

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

When computing roots numerically, the usual form of the quadratic formula is not always ideal, due to possible loss of significance. An alternative form is given by

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

The "$\backslash mp$" sign above indicates that the $x\_+\; \backslash$ root (as defined above) is calculated with the minus sign and the $x\_-\; \backslash$ root with the plus sign, rather than vice versa - see plus-minus sign for more detail.

This form may be useful in numerical analysis when high precision of the roots is required, particularly when a is large and the roots are very close together. However, this form imposes the additional requirement that c also be nonzero. If c were zero, the alternative formula will correctly give zero as a root, but will fail to give the non-zero root because it specifies division of zero by zero. Note also that the signs distinguishing the two roots are reversed.

Discriminant

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In the above formula, the term

$\backslash Delta\; =\; b^2\; -\; 4ac\backslash ,\backslash !$

is the discriminant of the quadratic equation, so called because it discriminates between three qualitatively different cases:

• If the discriminant is zero, there is a repeated solution x, and this solution is real. Geometrically, this means that the parabola described by the quadratic equation touches the x-axis in a single point. This is because when the discriminant is equal to zero, the term with the plus/minus sign vanishes and a single root remains. Sometimes this is called a double root and its value is:

$x\; =\; -\backslash frac\{b\}\{2a\}\; \backslash$

A discriminant of zero means the quadratic can be factorised to a perfect square.

• If the discriminant is positive, there are two different solutions x, both of which are real. Geometrically, this means that the parabola intersects the x-axis in two points. Furthermore, if the discriminant is a perfect square, the roots are rational numbers—in other cases they may be quadratic irrationals. The two real roots are:

$x\_+\; =\; \backslash frac\{-b\; +\; \backslash sqrt\; \{b^2-4ac\; \backslash \; \}\}\{2a\}\; \backslash$

$x\_-\; =\; \backslash frac\{-b\; -\; \backslash sqrt\; \{b^2-4ac\; \backslash \; \}\}\{2a\}\; \backslash$

If the discriminant is also a perfect square then the quadratic equation can be factorised in to two brackets.

• If the discriminant is negative, there are two different solutions x, both of which are complex numbers. The two solutions are complex conjugates of each other. In this case, the parabola does not intersect the x-axis at all. After factoring out −1 from the discriminant (or $i\; =\; \backslash sqrt\{-1\}$ from the square root radical), the two complex conjugate roots are:

$x\_+\; =\; \backslash frac\{-b\; +\; i\; \backslash sqrt\; \{4ac\; -\; b^2\; \backslash \; \}\}\{2a\}\; \backslash$

$x\_-\; =\; \backslash frac\{-b\; -\; i\; \backslash sqrt\; \{4ac\; -\; b^2\; \backslash \; \}\}\{2a\}\; \backslash$

A discriminant of less than zero means the quadratic can not be factorised.

Generalizations

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The formula and its proof remain correct if the coefficients $a\backslash ,\backslash !$, $b\backslash ,\backslash !$ and $c\backslash ,\backslash !$ are complex numbers, or more generally members of any field whose characteristic is not 2. (In a field of characteristic 2, the element 2a is zero and it is impossible to divide by it.)

The symbol

$\backslash pm\; \backslash sqrt\; \{b^2-4ac\}$

in the formula should be understood as "either of the two elements whose square is $b^2\; -\; 4ac\backslash ,\backslash !$, if such elements exist". In some fields, some elements have no square roots and some have two; only zero has just one square root, except in fields of characteristic 2. Note that even if a field does not contain a square root of some number, there is always a quadratic extension field which does, so the quadratic formula will always make sense as a formula in that extension field.

Characteristic 2

In a field of characteristic 2, the quadratic formula, which relies on 2 being a unit, does not hold. Consider the monic quadratic polynomial x² + bx + c over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is x = √c and note that there is only one root since –√c = –√c + 2√c = √c. In summary, x² + c = (x + √c)². Confer quadratic residue for more information about extracting square roots in finite fields.

In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they cannot be expressed in terms of square roots of numbers in the coefficient field. Instead, define the 2-root R(c) of c to be a root of the polynomial x² + x + c, an element of the splitting field of that polynomial. One verifies that R(c) + 1 is also a root. In terms of the 2-root operation, the two roots of the (non-monic) quadratic ax² + bx + c are

$\backslash frac\{b\}\{a\}R\backslash left(\backslash frac\{ac\}\{b^2\}\backslash right)$

and

$\backslash frac\{b\}\{a\}\backslash left(R\backslash left(\backslash frac\{ac\}\{b^2\}\backslash right)+1\backslash right).$

For example, let a denote a multiplicative generator of the group of units of F₄, the Galois field of order four (thus a and a + 1 are roots of x² + x + 1 over F₄). Because (a + 1)² = a, a + 1 is the unique solution of the quadratic equation x² + a = 0. On the other hand, the polynomial x + ax + 1 is irreducible over F₄, but splits over F₁₆, where it has the two roots ab and ab + a, where b is a root of x² + x + a in F₁₆.

Viète's formulas

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Viète's formulas give a simple relation between the roots of a polynomial and its coefficients. In the case of quadratic polynomial, they take the following form:

$x\_+\; +\; x\_-\; =\; -\backslash frac\{b\}\{a\}$

and

$x\_+\; \backslash cdot\; x\_-\; =\; \backslash frac\{c\}\{a\}.$

The first formula above yields a convenient expression when graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, when there are two real roots the vertex’s x-coordinate is located at the average of the roots (or intercepts). Thus the x-coordinate of the vertex is given by the expression:

$x\_V\; =\; \backslash frac\; \{x\_+\; +\; x\_-\}\; \{2\}\; =\; -\backslash frac\{b\}\{2a\}.$

The y-coordinate can be obtained by substituting the above result into the given quadratic equation, giving

$y\_V\; =\; -\; \backslash frac\{b^2\}\{4a\}\; +\; c\; =\; -\; \backslash frac\{\; \backslash Delta\}\; \{4a\}.$

Solving equations of a higher degree

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Certain higher-degree equations may be quadratic in form, such as:

$2x^6\; +\; 3x^3\; +\; 5\; =\; 0,\backslash ,$

which can be written

$2u^2\; +\; 3u\; +\; 5\; =\; 0\; \backslash$

where

$u\; =\; x^3\; \backslash$.

Note that the highest exponent is twice the value of the exponent of the middle term. This equation may be resolved directly or with a simple substitution, using the methods that are available for the quadratic, such as factoring (also called factorising), the quadratic formula, or completing the square.

History

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On clay tablets dated between 1800 BC and 1600 BC, the ancient Babylonians first discovered quadratic equations and also gave early methods for solving them. Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic equations of the form ax² = c and ax² + bx = c and also gave methods for solving them.

Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula. Euclid produced a more abstract geometrical method around 300 BC. The Bakshali Manuscript written in India between 200 BC and 400 CE introduced the general algebraic formula for solving quadratic equations, and also introduced quadratic indeterminate equations (origin of type ax/c = y).

The first mathematician to have found negative solutions with the general algebraic formula, was Brahmagupta (India, 7th century). (Persia, 9th century) developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) introduced the complete solution to Europe in his book Liber embadorum in the 12th century. Bhaskara II (India, 12th century) solved quadratic equations with more than one unknown.

Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation. His original work is lost but Bhaskara II later quotes Shridhara's rule:

Multiply both sides of the equation by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the coefficient of the unknown; then take the square root. ^*

Derivation

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The quadratic formula is derived by the method of completing the square.

$ax^2+bx+c=0\; \backslash ,\backslash !$

Dividing our quadratic equation by $a\backslash ,\backslash !$ (which is allowed because $a\backslash ,\backslash !$ is non-zero), we have

$x^2\; +\; \backslash frac\{b\}\{a\}\; x\; +\; \backslash frac\{c\}\{a\}=0$

which is equivalent to

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

The equation is now in a form in which we can conveniently complete the square. To "complete the square" is to add a constant (i.e., in this case, a quantity that does not depend on $x\backslash ,\backslash !$) to the expression to the left of "$=\backslash ,\backslash !$", that will make it a perfect square trinomial of the form $x^2+2xy+y^2\backslash ,\backslash !$. Since $2xy\backslash ,\backslash !$ in this case is $\backslash frac\{b\}\{a\}\; x$, we must have $y\; =\; \backslash frac\{b\}\{2a\}$, so we add the square of $\backslash frac\{b\}\{2a\}$ to both sides, getting

$x^2+\backslash frac\{b\}\{a\}x+\backslash frac\{b^2\}\{4a^2\}=-\backslash frac\{c\}\{a\}+\backslash frac\{b^2\}\{4a^2\}.$

The left side is now a perfect square; it is the square of $\backslash left(x\; +\; \backslash frac\{b\}\{2a\}\backslash right)$. The right side can be written as a single fraction; the common denominator is $4a^2\backslash ,\backslash !$. We get

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

Taking square roots of both sides yields

$\backslash left|x+\backslash frac\{b\}\{2a\}\backslash right|\; =\; \backslash frac\{\backslash sqrt\{b^2-4ac\backslash \; \}\}\{|2a|\}\backslash Leftrightarrow$$x+\backslash frac\{b\}\{2a\}=\backslash pm\backslash frac\{\backslash sqrt\{b^2-4ac\backslash \; \}\}\{2a\}$

Subtracting $\backslash frac\{b\}\{2a\}$ from both sides, we get

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

Derivation of the alternative expression

The alternative expression for the quadratic formula results from multiplying the top and bottom expression above with the conjugate of the numerator:

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

= \frac{\left ( -b \pm \sqrt {b^2-4ac\ } \right ) \left ( -b \mp \sqrt {b^2-4ac\ } \right )}{2a \left ( -b \mp \sqrt {b^2-4ac\ } \right )} = \frac{4ac}{2a \left ( -b \mp \sqrt {b^2-4ac} \right ) }

$x\_\backslash pm=\backslash frac\{2c\}\{-b\; \backslash mp\; \backslash sqrt\; \{b^2-4ac\backslash \; \}\}$

See also

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• Linear equation
• Cubic equation
• Quartic equation
• Quintic equation
• Fundamental theorem of algebra
• Parabola

External links

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• Quadratic Equation Solver
• Solve Quadratic equations, see work shown and draw graphs
• 101 uses of a quadratic equation part I Part II
• Quadratic graphical explorer Interactive applet. Sliders for a,b,c show effects on a graph.

Elementary algebra | Equations

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