From a course of school mathematics many remember that the root is a solution of the equation, that is those values X at which equality of its parts is reached. As a rule, the task of finding of the module of a difference of roots is set concerning quadratic equations, they can have two roots which difference you will be able to calculate.
Instruction
1. For a start solve the equation, that is find its roots or prove that they are absent. Before you the equation of the second degree: look whether it has AX2 appearance + BX + C = 0 where And, In and With – prime numbers and But it is not equal to 0.
2. If the equation is not equal to zero or at the second part of equality there is an unknown H, lead it to a standard look. For this purpose transfer all numbers to the left part, having replaced the sign facing them. For example, 2Х^2 + 3X + 2 = (-2X). It is possible to give this equation as follows: 2Х^2 + (3X + 2X) + 2 = 0. Now, when your equation is brought to a standard look, it is possible to start finding of its roots.
3. Calculate a discriminant of the equation of D. It is equal to the difference of B squared, and And, increased on With, and on 4. The equation cited as an example 2Х^2 + 5X + 2 = 0 has two roots as its discriminant is equal 5^2 + 4 x 2 x 2 = 9, that is it is more than 0. If the discriminant is equal to zero, you will be able to solve the equation, but it to have only one root. The negative discriminant demonstrates lack of roots of the equation.
4. Find a root from a discriminant (√D). For this purpose you can use the calculator with algebraic functions, online kulkulyatorom or the special table of roots (usually it is provided in the end of textbooks and reference books on algebra). In our case √D = √9 = 3.
5. To calculate the first root of a quadratic equation (X1), substitute in expression (-In + √D) the received number and divide result on And, increased by 2. That is X1 = (-5 + 3) / (2 x 2) =-0.5.
6. It is possible to find the second root of a quadratic equation X2 having replaced in a formula the sum with a difference, that is X2 = (-In - √D) / 2A. In the given example of X2 = (-5 - 3) / (2 x 2) =-2.
7. Take away from the first root of the equation of the second, that is X1 – X2. At the same time in what order you will substitute roots does not matter at all: the end result will be the same. The received number is the difference of roots, and you needed only to find the module of this number. In our case X1 – X2 =-0.5 - (-2) = 1.5 or H2 – X1 = (-2) - (-0.5) =-1.5.
8. The module is the distance on an axis of coordinates from zero to N point measured in single pieces therefore the module of any number cannot be negative. It is possible to find the number module as follows: the module of positive number is equal to him, and the module of negative – number opposite to it. That is |1.5| = 1.5 and |-1.5| = 1.5.