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Therefore, the solution = is extraneous and not valid, and the original equation has no solution. For this specific example, it could be recognized that (for the value x = − 2 {\displaystyle x=-2} ), the operation of multiplying by ( x − 2 ) ( x + 2 ) {\displaystyle (x-2)(x+2)} would be a multiplication by zero.
An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
One solution of the quadratic yields the user's location. The other solution is either ambiguous or extraneous – both can occur (which one depends upon the dimensions and the user location). Generally, eliminating the incorrect solution is not difficult for a human, but may require vehicle motion and/or information from another system.
The worked-example effect is a learning effect predicted by cognitive load theory. [1] [full citation needed] Specifically, it refers to improved learning observed when worked examples are used as part of instruction, compared to other instructional techniques such as problem-solving [2] [page needed] and discovery learning.
where x = 7 is the solution to the problem, and x = -4 is an extraneous solution because it is not pertinent to the problem. Tparameter 17:46, 19 January 2008 (UTC) No, this is not a sutiable example of an extraneous solution. Since x = -4 can satisfly the equation x 2 - 3x + 5 = 0, only does not satisfly the domain that sets manually.
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For example, the general quadratic equation is usually written ax 2 + bx + c = 0. The process of finding the solutions, or, in case of parameters, expressing the unknowns in terms of the parameters, is called solving the equation. Such expressions of the solutions in terms of the parameters are also called solutions.
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