So, than the irrational equation differs from rational? If the unknown variable to be under the sign of a square root, then the equation is considered irrational.
Instruction
1. The main method of the solution of such equations - a method of construction of both members of equation in a square. However. it is natural, first of all it is necessary to get rid of the sign of a square root. Technically this method is not difficult, but it can sometimes lead to troubles. For example, equation of v(2х-5) =v(4х-7). Having squared both of its parties, you receive 2kh-5=4kh-7. Such equation cannot be solved; x =1. But number 1 will not be a root of this equation. Why? Substitute unit in the equation instead of value x. Both in right and the left part will contain the expressions which are not making sense that is negative. Such value is not admissible for a square root. Therefore 1 - a foreign root and therefore this irrational equation has no roots.
2. So, the irrational equation is solved by means of a method of squaring of both of its parts. And having solved the equation, it is necessary to make surely check to cut foreign roots. For this purpose substitute the found roots in the original equation.
3. Review one more example. the 2nd +vkh-3=0konechno, it is possible to solve this equation according to the same scheme, as previous. To transfer the compound equations which do not have a square root to the right part and further to use a method of squaring. to solve the received rational equation and to check roots. But there is also other way, more graceful. Enter a new variable; vx =y. Respectively, you receive the look equation 2y2+y-3=0. That is usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Further solve two equations of vx =1; vx =-3/2. The second equation of roots has no, from the first we find that x =1. Do not forget, about need of check of roots.