Skills of the solution of the equations with degrees are required from pupils of all educational organizations, whether it be school, higher education institution or college. It is necessary to solve the sedate equations as in itself, and for the solution of other tasks (physical, chemical). To learn to solve such equations quite simply, the main thing to consider a number of small subtleties and to observe an algorithm.

## It is required to you

- Calculator

## Instruction

1. At first it is necessary to define to what look the available sedate equation belongs. It can be square, biquadratic or the equation with odd **degrees**. It is important to look at the highest degree. If it the second - that quadratic equation, if the first - linear. If the highest degree of the equation - the fourth, and is available a variable in the second degree and coefficient, then the equation further - biquadratic.

2. If in the equation there are two composed: the variable in any degree and coefficient, that equation is solved very simply: we transfer a variable to one member of equation, and number in another. Further we take a root of such degree from number in what there is a variable. If degree odd, you can write down the answer if even, decisions two - the considered number, and the counted number with the opposite sign.

3. It is quite simple to solve a quadratic equation too. The quadratic equation is the equation of a look: a*x^2+b*x+c=0. At first we consider an equation discriminant on a formula: D=b*b-4*a*c. Further everything depends on the sign of a discriminant. If the discriminant is less than zero, then we have no decisions. If the discriminant is more or is equal to zero, then we consider equation roots on a formula x= (-b-root (D)) / (2*a).

4. Biquadratic equation of type: a*x^4+b*x^2+c=0 is solved also quickly, as well as the previous two types of the sedate equations. For this purpose we use replacement of x^2=y, and we solve the biquadratic equation as square. We will receive as a result two y and we will pass back to x^2. That is, we will receive two equations of a type of x^2=a. How to solve such equation it was mentioned above.