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7.1 Solving equations. ... the term 2x in x 2 + 2x + 1 is a linear term in a quadratic polynomial. ... = 1/14 (x + 4)(x + 1)(x − 1)(x − 3) + 0.5.
Microsoft Math contains features that are designed to assist in solving mathematics, science, and tech-related problems, as well as to educate the user. The application features such tools as a graphing calculator and a unit converter. It also includes a triangle solver and an equation solver that provides step-by-step solutions to each problem.
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 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 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 variable y to be the unknown, and then the equation is solved by y = x – 1.
The solutions –1 and 2 of the polynomial equation x 2 – x + 2 = 0 are the points where the graph of the quadratic function y = x 2 – x + 2 cuts the x-axis. In general, an algebraic equation or polynomial equation is an equation of the form =, or = [a]
An example of multiplying binomials is (2x+1)×(x+2) and the first step the student would take is set up two positive x tiles and one positive unit tile to represent the length of a rectangle and then one would take one positive x tile and two positive unit tiles to represent the width. These two lines of tiles would create a space that looks ...
To begin solving, we multiply each side of the equation by the least common denominator of all the fractions contained in the equation. In this case, the least common denominator is ( x − 2 ) ( x + 2 ) {\displaystyle (x-2)(x+2)} .