How to find the root equation

If there are two values, and between them is a sign of equality, this is an example called by the equation. Harding an unknown, we will know the root. To declassify this unknown, you will have to work on the calculation.

It will be clearer if we take a specific equation: x + 10 \u003d 16-2x. It belongs to linear, make up its free members and unknown x. We are broaderring these components in different directions on the sign of equality. Now the equation has acquired such a type: 2x + x \u003d 16 - 10 or 3x \u003d 6; x \u003d 2. Result: x \u003d 2.

A little greater knowledge margin is needed to calculate the root in the example where the desired squared. This equation is square and the difference from linear in the fact that the results may be 1 or 2 or it will be found that the roots 0. To understand better, solve the equation: X, erected into the square, multiply by 3 + 3x \u003d 90. Make so that On the right was formed 0: x2 x 3 + 3x -90 \u003d 0. The numbers in front of x - coefficients 1, 3, 3. The determination of the discriminant is required: we are taken into a square 3 - the second coefficient and take the work of 1 and 3. As a result, we get 6 - it means By bringing to the end, it will be found that this equation of roots 2. If the discriminant was narrowed by a negative number, then it would be irrational to be sophisticated in the calculation of the roots - they simply do not. In case D \u003d 0, the root is only 1. Now I will fulfill the calculation to determine these 2 roots. To count 1 root to the second coefficient with a sign - add the root from D and divide it to a double first coefficient: -3 + square root of 16, divide on 2. will be released 1/2. The calculation of the second is similar, only the root of D is deducted. We are as a result - 3 whole and 1/2.

More difficult cubic square equation. The view of him is such: x3-3x2-4x + 20 \u003d 0. We select a number that can be divided into a free term so that 0. Dividers for 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20. It turns out that this is divider 5, it is also one of The desired roots. It remains to solve the square equation and all the roots are known.