Instruction

1

With the options can be equations and inequalities. In either case, we need to Express x.

Simply in such type of examples, this is not done explicitly, but using this parameter.

By itself, a parameter, more precisely, its value is a number. Settings are usually designated by the letter a. But the problem is that we don't know his module, no mark. Hence difficulties arise when working with inequalities or solving modules.

Simply in such type of examples, this is not done explicitly, but using this parameter.

By itself, a parameter, more precisely, its value is a number. Settings are usually designated by the letter a. But the problem is that we don't know his module, no mark. Hence difficulties arise when working with inequalities or solving modules.

2

However, it is possible (but beware, pre-noting all possible restrictions) apply all the usual methods of working with equations and inequalities.

And, in principle, the expression x through and usually takes a lot of time and effort.

But writing the full answer is much more laborious and time-consuming process.

And, in principle, the expression x through and usually takes a lot of time and effort.

But writing the full answer is much more laborious and time-consuming process.

3

The fact that, in connection with ignorance of the value of the parameter, we must consider all possible cases for all values of a from minus to plus infinity.

Here we can use graphical method. Sometimes called "coloring". It lies in the fact that we are in x(a) (or a(x) as the image represented by lines, the resulting transformation of our original example. And then we begin to work with these lines: since the a value is not fixed, we need the line containing in its equation a parameter shift on schedule, in parallel to tracking and calculating points of intersection with other lines, as well as analyzing the signs of the areas they are suitable for us or not. Suitable for convenience and clarity, let's shade.

Thus, we pass the entire number line from minus to plus infinity, checking the answer for all.

Here we can use graphical method. Sometimes called "coloring". It lies in the fact that we are in x(a) (or a(x) as the image represented by lines, the resulting transformation of our original example. And then we begin to work with these lines: since the a value is not fixed, we need the line containing in its equation a parameter shift on schedule, in parallel to tracking and calculating points of intersection with other lines, as well as analyzing the signs of the areas they are suitable for us or not. Suitable for convenience and clarity, let's shade.

Thus, we pass the entire number line from minus to plus infinity, checking the answer for all.

4

The written response similar to the response for the method of intervals with a certain caveat: we don't just define a set of solutions for x, and write some set of values and corresponds to how many values of X.

# Advice 2: How to solve equations with parameters

When solving problems with

**parameters**it is important to understand the condition. Solve the equation with the – a option means to write the answer to any of the possible values of the parameter. The answer should reflect the enumeration of the whole number line.

Instruction

1

The simplest type of tasks with parameters – problem on square trinomial A·x2+B·x+C Parametric value can be any of the coefficients

**of the equation**: A, B or C. to Find the roots of trinomial square for every parameter value – so to solve the quadratic equation A·x2+B·x+C=0, listing each of the possible values of the unfixed variables.2

In principle, if the equation A·x2+B·x+C=0 is a parameter of the senior coefficient A, it will be a square only when A≠0. When A=0 it degenerates into the linear equation B·x+C=0 has one root: x=-C/B. Therefore, test condition A≠0, A=0 must be the first line item.

3

The quadratic equation has real roots with non-negative discriminant D=B2-4·A·C If D>0 it has two distinct roots, if D=0 only one. Finally, if D

4

Often for solving problems with the settings, use the theorem of vieta. If the quadratic equation A·x2+B·x+C=0 has roots x1 and x2, the true system: x1+x2=-B/A x1·x2=C/A. the quadratic equation leading coefficients equal to one, we see that: x2+M·x+N=0. For him, the vieta theorem has a simplified form: x1+x2=-M, x1·x2=N. it Should be noted that the vieta theorem is true in the presence of one or two roots.

5

The same roots are found by using the vieta theorem, we can substitute back into the recording equation: x2-(x1+x2)·x+x1·x2=0. Do not confuse: here x is a variable, x1, and x2 - specific number.

6

Often helps in deciding the method of factorization. Let the equation A·x2+B·x+C=0 has roots x1 and x2. When the true identity of A·x2+B·x+C=A·(x-x1)·(x-x2). If the root only, then you can just say that x1=x2, then A·x2+B·x+C=A·(x-x1)2.

7

**Example.**Find all integers p and q, in which the roots of the equation x2+p·+q=0 is equal to p and q.

**Solution.**Let p and q satisfy the conditions of the problem, that is, they are roots. Then by vieta theorem:p+q=-p,pq=q.

8

The system is equivalent to the conjunction p=0, q=0 or p=1, q=-2. It now remains to check is to make sure that the resulting numbers do satisfy the problem statement. For this you just need to substitute the numbers into the original equation.

**Answer:**p=0, q=0 or p=1, q=-2.