How To Find The Area Of ​​Feasible Solutions | The Science

The Science

# How to find the area of ​​feasible solutions

Once the roots of the equation are found, make sure that after the substitution equality will make sense. And if the substitution is very complex, and the roots of a large number of the most rational way to answer this question is to find the area of ​​"acceptable solutions", which separates the suitable options.

## Instruction how to find the area of ​​feasible solutions

Step 1:

Determine whether the problem of the physical meaning. So, if the problem of determining the area is reduced to a quadratic equation, it is evident that the negative space can not be: the range of permissible values ​​[0; infinity). If you are in solving got a couple of roots -3, 3, it is obvious that in TCC -3 misses.

Step 2:

Decide whether you need complex values. The use of such allows you to remove restrictions from the values ​​of the trigonometric functions, the numbers "under the root" and a number of other situations. Students this item can be safely ignored, since Even USE presence of complex numbers ignores.

Step 3:

Consider your expression and define the "state" of unknown variables. Are these arguments, a function (sin (x))? Are they in the numerator or the denominator? Elevated to a whole, fractional or negative power? Consider all the variables in this (obviously, it may occur in several locations of the equation).

Step 4:

Remember, any restrictions imposed on each function variable. For example, it is known that, in general, the denominator can not be zero. Therefore, if the bottom of the fractions formed two x-function, then it falls from TCC x = 2, since this violates the meaning of the equation. A simpler example: under the root can only be positive. Therefore, if you come across design "under the root of x", then you can safely limit the DHS as a variable x [0, infinity).

Step 5:

Draw a number line and move it all the restrictions imposed by example. The "forbidden" zones of shade, some points appear as hollow circles. Once everything is done, "empty" area will significantly direct DHS equal: if the solution gets into the equation without hatch segment, the answer is acceptable. If these areas are not left, the above example does not have solutions.