Quadratic equation
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In mathematics, a quadratic equation is a polynomial equation of the second degree. The generalized form is
- <math>ax^2+bx+c=0\mbox{ , where }a\ne 0. \,</math>
The letters <math>a\,\!</math>, <math>b\,\!</math> and <math>c\,\!</math> are called coefficients: <math>a\,\!</math> is the coefficient of <math>x^2\,\!</math>, <math>b\,\!</math> is the coefficient of <math>x\,\!</math>, and <math>c\,\!</math> is the constant coefficient, also called the free term.
A quadratic equation with real or complex coefficients has two complex roots (i.e., solutions for <math>x\,\!</math> when <math>y = ax^2+bx+c\mbox{ and }y = 0\,\!</math>) usually denoted as <math>x_1\,\!</math> and <math>x_2\,\!</math>, although the two roots may be equal. These roots can be computed using the quadratic formula.
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Quadratic formula
The quadratic formula is an explicit definition of (and an algorithm for) the solutions of a quadratic equation in terms of the coefficients a, b, and c, which we temporarily assume to be real (but see below for generalizations), and with a being non-zero. These solutions are also called the roots of the equation. The formula reads
- <math>x_{1,2} = \frac{-b \pm \sqrt {b^2-4ac \ }}{2a} \ </math>
Template:ExampleSidebar{2(8)}</math> This gives the solutions <math>x_1 = \frac{3}{2}</math> and <math>x_2 = -\frac{11}{4}</math>.}}
The term <math>\Delta = b^2 - 4ac\,\!</math> is called the discriminant of the quadratic equation, 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:
- <math>x = -\frac{b}{2a} \ </math>
- 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:
- <math>x_1 = \frac{-b + \sqrt {b^2-4ac \ }}{2a} \ </math>
- <math>x_2 = \frac{-b - \sqrt {b^2-4ac \ }}{2a} \ </math>
- 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 <math> i = \sqrt{-1} </math> from the square root radical), the two complex conjugate roots are:
- <math>x_1 = \frac{-b + i \sqrt {4ac - b^2 \ }}{2a} \ </math>
- <math>x_2 = \frac{-b - i \sqrt {4ac - b^2 \ }}{2a} \ </math>
When computing roots numerically, the usual form of the quadratic formula is not ideal, due to possible loss of significance. An alternative form is given by
- <math>x_{1,2}=\frac{2c}{-b \mp \sqrt {b^2-4ac\ }}</math>
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.
Derivation
The quadratic formula is derived by the method of completing the square.
- <math>ax^2+bx+c=0 \,\!</math>
Dividing our quadratic equation by <math>a\,\!</math> (which is allowed because <math>a\,\!</math> is non-zero), we have
- <math>x^2 + \frac{b}{a} x + \frac{c}{a}=0</math>
which is equivalent to
- <math>x^2+\frac{b}{a}x=-\frac{c}{a}.</math>
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 <math>x\,\!</math>) to the expression to the left of "<math>=\,\!</math>", that will make it a perfect square trinomial of the form <math>x^2+2xy+y^2\,\!</math>. Since <math>2xy\,\!</math> in this case is <math>\frac{b}{a} x </math>, we must have <math>y = \frac{b}{2a}</math>, so we add the square of <math>\frac{b}{2a}</math> to both sides, getting
- <math>x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}.</math>
The left side is now a perfect square; it is the square of <math>\left(x + \frac{b}{2a}\right)</math>. The right side can be written as a single fraction; the common denominator is <math>4a^2\,\!</math>. We get
- <math>\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.</math>
Taking square roots of both sides yields
- <math>\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac\ }}{|2a|}\Leftrightarrow</math><math>x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}</math>
Subtracting <math>\frac{b}{2a}</math> from both sides, we get
- <math>x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac\ }}{2a}=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}.</math>
Alternative expression
The alternative expression for the quadratic formula results from multiplying the top and bottom expression above with the conjugate of the numerator:
- <math>x_{1,2}= \frac{-b \pm \sqrt {b^2-4ac\ }}{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 ) } </math>
- <math>x_{1,2}=\frac{2c}{-b \pm \sqrt {b^2-4ac\ }}</math>
Generalizations
The formula and its proof remain correct if the coefficients <math>a\,\!</math>, <math>b\,\!</math> and <math>c\,\!</math> are complex numbers, or more generally members of any field whose characteristic is not <math>2\,\!</math>. (In a field of characteristic <math>2\,\!</math>, the element <math>2a\,\!</math> is zero and it is impossible to divide by it.)
The symbol
- <math>\pm \sqrt {b^2-4ac }</math>
in the formula should be understood as "either of the two elements whose square is <math>b^2 - 4ac\,\!</math>, 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 <math>2\,\!</math>.
Viète's formulas
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:
- <math> x_1 + x_2 = -\frac{b}{a} </math>
- <math> x_1 \cdot x_2 = \frac{c}{a}</math>
This yields a convenient expression when graphing a quadratic function. Since the graph is symmetric vertically about 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:
- <math> x_V = \frac {x_1 + x_2} {2} = \frac {\frac{-b + \sqrt {b^2-4ac\ }}{2a} + \frac{-b - \sqrt {b^2-4ac\ }}{2a}}{2} = -\frac{b}{2a} </math>
The y-coordinate can be obtained by substituting the above result into the function.
- <math> y_V = - \frac{ \Delta} {4a} </math>
<math> V = (x_V,y_V) = \left( -\frac{b}{2a}, - \frac{ \Delta} {4a} \right) </math>
Solving Equations of a Higher Degree
Certain higher-degree equations may be quadratic in form, such as:
- <math>2x^6+3x^3+5=0 \,\!</math>.
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
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). Muḥammad ibn Mūsā al-Ḵwārizmī (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. [1]
See also
- linear equation
- cubic equation
- quartic equation
- quintic equation
- fundamental theorem of algebra
- parabola
External links
- Quadratic Equation Solver
- Solve Quadratic equations, see work shown and draw graphs
- 101 uses of a quadratic equation part I Part IIcs:Kvadratická rovnice
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