# Equation Solver

You should be able to solve quadratic equations and understand the full complement of squares. As the students can see how are the equations of the third and fourth degrees. It’s not particularly important, but entertaining. Here I must say that the most common approach equation solver with steps is to use the formulas Cardano and Ferrari’s method, which can seem complex. There are however quite basic approaches to the solution of equations of the third and fourth degrees through the usual variable substitution and the introduction of additional unknown to fit the formula under view square. Such approaches have several deficiencies and does not reveal the properties of polynomials, but they are pretty simple to understand at the school level. Equation solver wolfram methods may be a useful exercise.

You should be able to solve the main types of private cases, equations of higher degrees. For example, the equations of the type ax $latex ^ {2n} + bc ^ n + c = 0 or symmetric equation coefficients.

Immediately it is worth mentioning that the analytic equations over the fourth degree generally hesitant is the assertion is remarkable theorem of Abel. The average reader it makes no sense to go into evidence, but interested in theoretical mathematics, I highly recommend that you become familiar with the book “Abel’s theorem in problems and solutions».

Need to know the basic theorem of algebra. It is that any polynomial is a complex root. At this stage, to understand equation solver with variables of this theorem you will fail, but the basic intuition on graphs of complex functions can tell why the theorem is true. To do this, you can consider the image Freehand circle in the complex plane for an arbitrary polynomial. This is a kind of closed line. The origin proves to be either inside or outside this line. In the first case, you can reduce the radius of the primary circle until the line crosses the origin (the intersection of the and root), the second on the contrary increase the intersection RADIUS as the origin of the image. This evidence very many holes, but while you are unlikely to be able to articulate something more severe. However, this illustration gives a good insight and understanding, the same strictness comes later.

The possibility of divisibility of polynomials (with and without balance) and from the fundamental theorem of algebra must be clear that any nth degree polynomial will be presenting in the form of $latex a\prod_ {i = 0} ^ n (x-x_i) $, where $latex x_i $ — the complex roots of a polynomial. From here simply displays the formula Viêt in general terms (if you brought it yourself, the binomial theorem and formula Viêt hence disable). If we know that all roots of the polynomial are rational, then the formulas should be Viêt obvious way when they attack.

If you know that a complex $ x_0 $latex is a root of the polynomial $latex f $ with real coefficients, then substituting into him coupled $latex \bar{x_0} $, we make sure that it is the root. From here you can get a similar view above the real polynomial in the form of $latex a\prod_ {i = 0} ^ n (x-x_i) \prod_{j = 0} ^ m (x ^ 2 + p_jx + q_j) $, which uses only real roots and coefficients. It is now clear that any real polynomial of odd degree has a real root.

As an equation solver by steps, but a very important and interesting, I recommend at this point to learn about generating functions. They are almost nowhere to pass engineering degrees, but they are extremely important in modern science and have the broadest application. In math problem solver tasks of medium complexity and some related fields (analysis of sequences probability theory) is now one of the most solve equation algebraically of mathematical tools.

With inequalities all pretty simple — enough to understand how basic cases at the level of the books for the exam. There’s nothing complex. Similarly with systems of equations. It is important to understand the general principle of solving systems of equations through the expression first one of n variables through n-1 $latex $ other, then this value substitution system, an expression of one of the remaining through the rest, etc.

You can view the same way toward Gauss method solving systems of linear equations, but actually it is not yet necessarily — motivation for solving such systems and their significance will become apparent later.

Exercises:

a) decide to $latex equation x ^ 5 + 5 x ^ 4 + x ^ 3 + x ^ 2 + 5 x +1 = $0 less than two minutes.

b) Using formulas to get representation for Viêt discriminant of a quadratic equation solver in the form $latex (D) = a ^ 2 (x_1 x_2-) ^ 2, where $latex x_i $ — the roots of a polynomial. Unlike the usual school formulas, this definition generalizes to the case of polynomials of arbitrary order: $latex D = a_0 ^ {2n-2} \prod_{i < j} (x_i-x_j) ^ 2 $. Using this definition and formulas Viêt find discriminant for polynomial x ^ 3 $latex + px + q $.

any parameters) a $latex equation \sum_{i = 1} ^ x {1 \ over i (+1 i)} = a $ has the solution and what? (Worth considering only integer $latex x\ge $1).