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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 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.
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.
For example, the statement = is true if is either 2 or −2 and false otherwise. [26] Equations with variables can be divided into identity equations and conditional equations. Identity equations are true for all values that can be assigned to the variables, such as the equation 2 x + 5 x = 7 x {\displaystyle 2x+5x=7x} .
Given two different points (x 1, y 1) and (x 2, y 2), there is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2, the slope of the line is . Thus, a point-slope form is [3]
By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...
How to Solve It suggests the following steps when solving a mathematical problem: . First, you have to understand the problem. [2]After understanding, make a plan. [3]Carry out the plan.
This can be seen in the following tables, the left of which shows Newton's method applied to the above f(x) = x + x 4/3 and the right of which shows Newton's method applied to f(x) = x + x 2. The quadratic convergence in iteration shown on the right is illustrated by the orders of magnitude in the distance from the iterate to the true root (0,1 ...