Instruction

1

Linear equation in General case has the form: ax + b = 0, and the unknown variable x here can be only in the first degree, she also should not be in the denominator. However, the problem is often the equation appears, for example, like this: x+2/4 + x = 3 – 2*x. In this case, before computing the argument it is necessary to bring the equation to the common view. This is accomplished by a series of transformations.

2

Drag a second (right) part

**of the equation**on the other side of equality. Each term will change its sign: x+2/4 + x - 3 + 2*x = 0. Swipe the addition of arguments and numbers, simplify the expression: 4*x – 5/2 = 0. Thus, the obtained General form write linear**equations**, it is easy to find x: 4*x = 5/2, x = 5/8.3

In addition to the operations described above, when solving equations, you should use 1 and 2 the identity transformation. Their essence lies in the fact that both parts

**of the equation**can be folded with the same or multiply the same number or expression. The resulting equation will look different, but its roots will remain unchanged.4

The solution of quadratic equations of the form ах2 + bх +C = 0 reduces to determining the coefficients a, b, C and their substitution in known formulas. As a rule, to give a General view you must do the transformation and simplification of expressions. Thus, the equation of the form x2 = (6x + 8)/2 open the brackets, leaving the right of the equal sign. Get the following entry: -x2 - 3x + 4 = 0. Multiply both parts of the equality by -1 and note the result: x2 + 3x - 4 = 0.

5

Calculate the discriminant of the square

**equation**by the formula D = b2 – 4*a*c = 32 – 4*1*(-4) = 25. In case of positive discriminant the equation has two roots, formula of finding of which are: x1 = -b + √(D)/2*a; x2 = -b - √(D)/2*a. Substitute values and calculate: x1 = (-3+5)/2 = 1 x2 = (-3-5)/2 = -4. If the resulting discriminant is equal to zero, the equation would have only one root, it follows from the formulas for D6

When finding the roots of cubic equations using the method of vieta-Cardano. More complex

**equations**of degree 4 are calculated using the replacement, which decreases the degree of the arguments and**equations**are solved in several stages, as the square.