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A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating and , which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved. [14]
For instance, if the one solving the math word problem has a limited understanding of the language (English, Spanish, etc.) they are more likely to not understand what the problem is even asking. In Example 1 (above), if one does not comprehend the definition of the word "spent," they will misunderstand the entire purpose of the word problem.
The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
A quadratic equation is one which includes a term with an exponent of 2, for example, , [40] and no term with higher exponent. The name derives from the Latin quadrus , meaning square. [ 41 ] In general, a quadratic equation can be expressed in the form a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0} , [ 42 ] where a is not zero (if it were ...
Figure 1. Plots of quadratic function y = ax 2 + bx + c, varying each coefficient separately while the other coefficients are fixed (at values a = 1, b = 0, c = 0). A quadratic equation whose coefficients are real numbers can have either zero, one, or two distinct real-valued solutions, also called roots.
Given a quadratic polynomial of the form + + it is possible to factor out the coefficient a, and then complete the square for the resulting monic polynomial. Example: + + = [+ +] = [(+) +] = (+) + = (+) + This process of factoring out the coefficient a can further be simplified by only factorising it out of the first 2 terms.
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