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The quadratic formula
In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing. Using the quadratic formula is often the most convenient way.
The general quadratic equation is
{\displaystyle ax^{2}+bx+c=0.} ax^{2}+bx+c=0.
Here x represents an unknown, while a, b, and c are constants with a not equal to 0. One can verify that the quadratic formula satisfies the quadratic equation, by inserting the former into the latter. With the above parameterization, the quadratic formula is:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.} {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}
Each of the solutions given by the quadratic formula is called a root 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. As well as being a formula that will yield the zeros of any parabola, the quadratic equation will give the axis of symmetry of the parabola, and it can be used to immediately determine how many zeros it has.
Contents [hide]
1 Derivation of the formula
2 Geometrical significance
3 Historical development
4 Other derivations
4.1 Alternate method of completing the square
4.2 By substitution
4.3 By using algebraic identities
4.4 By Lagrange resolvents
4.5 By extrema
5 Dimensional analysis
6 See also
7 References
Derivation of the formula[edit]
The quadratic formula can be derived with a simple application of technique of completing the square.[1][2] For this reason, the derivation is sometimes left as an exercise for students, who can thus experience rediscovery of this important formula.[3][4] The explicit derivation is as follows.
Divide the quadratic equation by a, which is allowed because a is non-zero:
{\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0.} 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}}.} 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 can be applied. Thus, add a constant to both sides of the equation such that the left hand side becomes a complete square:
{\displaystyle x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+\left({\frac {b}{2a}}\right)^{2},} 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}}}.} \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 this:
{\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}.} \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}}.} x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac\ }}{2a}}.
Isolating x gives the quadratic formula:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.} x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.
The plus-minus symbol "±" indicates that both
{\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}} x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}
are solutions of the quadratic equation.[5] There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of a.
Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax2 − 2bx + c = 0[6] or ax2 + 2bx + c = 0,[7] where b has a magnitude one half of the more common one. These result in slightly different forms for the solution, but are otherwise equivalent.
A lesser known quadratic formula, as 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}}}}.} x={\frac {-2c}{b\pm {\sqrt {b^{2}-4ac}}}}.
In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing. Using the quadratic formula is often the most convenient way.
The general quadratic equation is
{\displaystyle ax^{2}+bx+c=0.} ax^{2}+bx+c=0.
Here x represents an unknown, while a, b, and c are constants with a not equal to 0. One can verify that the quadratic formula satisfies the quadratic equation, by inserting the former into the latter. With the above parameterization, the quadratic formula is:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.} {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}
Each of the solutions given by the quadratic formula is called a root 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. As well as being a formula that will yield the zeros of any parabola, the quadratic equation will give the axis of symmetry of the parabola, and it can be used to immediately determine how many zeros it has.
Contents [hide]
1 Derivation of the formula
2 Geometrical significance
3 Historical development
4 Other derivations
4.1 Alternate method of completing the square
4.2 By substitution
4.3 By using algebraic identities
4.4 By Lagrange resolvents
4.5 By extrema
5 Dimensional analysis
6 See also
7 References
Derivation of the formula[edit]
The quadratic formula can be derived with a simple application of technique of completing the square.[1][2] For this reason, the derivation is sometimes left as an exercise for students, who can thus experience rediscovery of this important formula.[3][4] The explicit derivation is as follows.
Divide the quadratic equation by a, which is allowed because a is non-zero:
{\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0.} 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}}.} 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 can be applied. Thus, add a constant to both sides of the equation such that the left hand side becomes a complete square:
{\displaystyle x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=-{\frac {c}{a}}+\left({\frac {b}{2a}}\right)^{2},} 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}}}.} \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 this:
{\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}.} \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}}.} x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac\ }}{2a}}.
Isolating x gives the quadratic formula:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.} x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.
The plus-minus symbol "±" indicates that both
{\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}} x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}
are solutions of the quadratic equation.[5] There are many alternatives of this derivation with minor differences, mostly concerning the manipulation of a.
Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax2 − 2bx + c = 0[6] or ax2 + 2bx + c = 0,[7] where b has a magnitude one half of the more common one. These result in slightly different forms for the solution, but are otherwise equivalent.
A lesser known quadratic formula, as 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}}}}.} x={\frac {-2c}{b\pm {\sqrt {b^{2}-4ac}}}}.