Increasing roots and losing roots: a logical problem of solving equations

The application of the equation reduces the amount of thinking that solves the arithmetic problem. With the equation, you don't have to cut your legs to get the chickens and rabbits to cage. An equation, or a set of equations, can be solved by algebraic transformation of the equation.

However, the algebraic transformation of the equation is logical and cannot be arbitrarily changed. A few days ago, I saw a question like this on Quora:

Known

\ (X+\frac{1}{x}=2\sqrt{5} \) (1)

Please

\ ({x}^{2}+\frac{1}{{x}^{2}} \) (2)

If it is solved directly from equation (1) (x\), then substituting (2), it is very troublesome. But if it's squared on the (1) side:

\ ({x}^{2}+\frac{1}{{x}^{2}}+2=20 \) (3)

Minus 2 on both sides, not hard to get.

\ ({x}^{2}+\frac{1}{{x}^{2}}=18 \) (4)

So the question is: is this reasonable? Any leaks? After someone has published this simple solution, it is immediately questioned, saying that this is not insurance, easy to lose root or generate root. For example \ (x=2 \) This obviously has and only one solution (real range) of the equation, put it on both sides of the square, instead get \ ({x}^{2}=4 \), there are two solutions.

Why is that? Two numbers are equal, their squares should be equal, and there is no doubt that there is more than one root.

This is because the squared on both sides of the equation is not equivalent to deformation. That is, \ (a=b \) can launch \ ({a}^{2}={b}^{2} \), \ ({a}^{2}={b}^{2} \) cannot be pushed out \ (a=b \). Only two equations can be introduced to each other, they are equivalent to each other deformation , in order to say that the two equations are essentially an equation, the solution is the same. When solving the equation, be sure to pay attention to this point.

So, put (1) square on both sides, not rigorous. However, this does not mean that we have to solve the x\. In turn, \ (t={x}^2+\frac{1}{{x}^{2}} \), you get:

\ (t+2={x}^2+\frac{1}{{x}^{2}}+2 \)

\ (t+2={(x+\frac{1}{x})}^{2} \)

\ (t+2=20 \)

\ (t = 18 \)

There is nothing logically wrong with this.

Increasing roots and losing roots: a logical problem of solving equations