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The roots of the quadratic function y = 1 / 2 x 2 − 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.
For instance, the square of the linear polynomial x + 1 is the quadratic polynomial (x + 1) 2 = x 2 + 2x + 1. One of the important properties of squaring, for numbers as well as in many other mathematical systems, is that (for all numbers x), the square of x is the same as the square of its additive inverse −x.
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.
The plus sign (+) and the minus sign (−) are mathematical symbols used to denote positive and negative functions, respectively. In addition, the symbol + represents the operation of addition, which results in a sum, while the symbol − represents subtraction, resulting in a difference. [1]
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, the power (+) expands into a polynomial with terms of the form , where the exponents and are nonnegative integers satisfying + = and the coefficient of each term is a specific positive integer ...
The names for the degrees may be applied to the polynomial or to its terms. For example, the term 2x in x 2 + 2x + 1 is a linear term in a quadratic polynomial. The polynomial 0, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero.
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...
Subtract the product just obtained from the appropriate terms of the original dividend (being careful that subtracting something having a minus sign is equivalent to adding something having a plus sign), and write the result underneath (x 3 − 2x 2) − (x 3 − 3x 2) = −2x 2 + 3x 2 = x 2 Then, "bring down" the next term from the dividend.