To solve roots, or the irrational equations, learn in the 8th class. As a rule, the main reception for finding of the decision in this case is the method of squaring.
Instruction
1. The irrational equations need to be given to rational to find the answer, having solved it in the traditional way. However except squaring one more action is added here: rejection of a foreign root. This concept is connected with irrationality of roots, i.e. this solution of the equation which substitution results in senselessness, for example, a root from a negative number.
2. Let's review the simplest example: √ (2 · x + 1) = 3. Square both parts of equality: 2 · x + 1 = 9 → x = 4.
3. It turns out that x=4 is a root at the same time and the usual equation 2•x + 1 = 9 and initial irrational √ (2•x + 1) = 3. Unfortunately, not always it happens simply. Sometimes the method of squaring results in absurdity, for example: √ (2 · x - 5) = √ (4•x - 7)
4. It would seem, it is necessary just to build both parts in the second degree and everything, a solution is found. However in reality the following turns out: 2 · x – 5 = 4•x – 7 →-2•x =-2 → x=1. Substitute the found root in the initial equation: √ (-3) = √ (-3) .x=1 is also called a foreign root of the irrational equation which has no other roots.
5. The example is more difficult: √ (2 · x \+ 5 · x - 2) = x – 6 ↑²2·x \+ 5 · x – 2 = x² – 12•x + 36x² + 17•x – 38 = 0
6. Solve a usual quadratic equation: D = 289 + 152 = 441x1 = (-17 + 21)/2 = 2; x2 = (-17 - 21)/2 =-19.
7. Substitute x1 and x2 in the initial equation to cut foreign roots: √ (2•2² + 5•2 - 2) = 2 – 6 → √16 =-4; √ (2 • (-19)² - 5•19 - 2) =-19 – 6 → √625 =-25. This decision the incorrect, therefore, equation, as well as previous, has no roots.
8. Example with replacement of a variable. Happens that simple construction of both members of equation in a square does not exempt from roots. In this case it is possible to use a replacement method: √ (x \+ 1) + √ (x² + 4) = 3 [y² = x² + 1] y + √ (y² + 3) = 3 → √ (y² + 3) = 3 – y ↑²
9. y² + 3 = 9 - 6 ·y + y²6•y = 6 → y=1.x² + 1 = 1 → x=0.
10. Check result: √ (0² + 1) + √ (0² + 4) = 1 + 2 = 3 – equality is observed, so root x=0 is the valid solution of the irrational equation.