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Quadratic formula. The roots of the quadratic function y = 1 2 x2 − 3x + 5 2 are the places where the graph intersects the x -axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
[1] [2] [3] [better source needed]. For example, 3 x 2 − 2 x y + c {\displaystyle 3x^{2}-2xy+c} is an algebraic expression. Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression:
Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
Algebraic fraction. In algebra, an algebraic fraction is a fraction whose numerator and denominator are algebraic expressions. Two examples of algebraic fractions are and . Algebraic fractions are subject to the same laws as arithmetic fractions. A rational fraction is an algebraic fraction whose numerator and denominator are both polynomials.
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
In mathematics, a quadratic form is a polynomial with terms all of degree two ("form" is another name for a homogeneous polynomial). For example, is a quadratic form in the variables x and y. The coefficients usually belong to a fixed field K, such as the real or complex numbers, and one speaks of a quadratic form over K.
Solving an equation symbolically means that expressions can be used for representing the solutions. For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2 (y + 1) – 1, a true statement. It is also possible to take the ...
is a horizontal line with y-intercept a0. The graph of a degree 1 polynomial (or linear function) f(x) = a0 + a1x, where a1 ≠ 0, is an oblique line with y-intercept a0 and slope a1. The graph of a degree 2 polynomial. f(x) = a0 + a1x + a2x2, where a2 ≠ 0. is a parabola. The graph of a degree 3 polynomial.