At the solution of tasks with parameters the main thing is to understand a condition. To solve the equation with parameter – means to write down the answer for any of possible values of parameter. The answer has to reflect search of all numerical straight line.
Instruction
1. The simplest type of tasks with parameters – tasks on square trinomial A·x \· x+C. Any of equation coefficients can become parametrical size: A, B or C. To find roots of a square trinomial – means to any of values of parameter to solve quadratic equation A·x \· x+C=0, having touched each of possible values of unstable size.
2. In principle, if in equation A·x \· x+C=0 is parameter the senior coefficient of A, it will be square only when A≠0. At A=0 it degenerates in the linear equation of B · x+C=0 having one root: x=-C/B. Therefore check of a condition of A≠0, A=0 has to go the first point.
3. The quadratic equation has the valid roots at a non-negative discriminant of D=B²-4·A·C. At D> 0 it has two various roots, at D=0 only one. At last, if D
4. Often Vieta theorem is applied to the solution of tasks with parameters. If quadratic equation A·x \· x+C=0 has roots x1 and x2, for them the system is right: x1+x2=-B/A, x1 · x2=C/A. The quadratic equation with the senior coefficient equal to unit, is called brought: x \· x+N=0. For it Vieta theorem has the simplified appearance: x1+x2=-M, x1 · x2=N. It should be noted that Vieta theorem is right in the presence both one, and two roots.
5. The same roots found by means of Vieta theorem can be substituted back in record of the equation: x \(x1+x2) · x+x1 · x2=0. Do not confuse: here x - a variable, x1 and x2 - concrete numbers.
6. Often helps at the decision a decomposition method on multipliers. Let equation A·x \· x+C=0 have roots x1 and x2. Then identity A·x \· is faithful to x+C=A · (x-x1) · (x-x2). If a root only, then it is possible just to tell that x1=x2, and then A·x \· x+C=A · (x-x1)².
7. Example. Find all numbers p and q at which roots of the equation x²+p·+q=0 are equal to p and q. Decision. Let p and q meet a statement of the problem, that is, are roots. Then on Vieta theorem: p+q=-p, pq=q.
8. The system is equivalent to set of p=0, q=0, or p=1, q=-2. Now it was necessary to make check - to make sure that the received numbers really meet a statement of the problem. For this purpose it is necessary just to substitute numbers in the initial equation. Answer: p=0, q=0 or p=1, q=-2.