How to prove this that every previous roots satisfy the newer ones if we make it more complex?

Question

We have $5-y^2=1$ ;it has 2 roots 2, -2.

Solving another way:

\begin{align} & 5-1×(y^2)=1 \\ \implies & 5-(5-y^2)y^2=1 \\ \implies & 5-5y^2 + y^4 =1 \\ \implies & y^4 -5y^2 +4×1=0 \\ \implies & y^4 -5y^2 +4×(5-y^2) =0 \\ \implies & y^4-9y^2 +20=0 \end{align} has roots two extra roots. Now, this is confusing as this thing can be done to any other random equation and if we keep on doing this we will be getting infinity roots. So for every equation there exist infinity roots which are hidden. Please guide me as I want to know where am I going wrong? Ok i understood that. Now if we go on doing this say if the 4 in power of y is multiplied by 1. And even the coefficients and we go on getting equations will 2,-2 always satisfy the new equation? To be more precise, will the roots of previous equation always satisfy the roots of newer equation? How do I prove it?

Answer 1

The mathematicians answered your question in their edits with the use of $\implies$ in place of your original "or".

What the symbol means is: If the previous statement is true (e.g. $5-1 \times y^2=1$), then so is the following statement ($5-(5-y^2)y^2=1$). In this case, you go from one statement to the next by a substitution (to be honest I'm not sure where others are getting the idea that you multiplied two equations?).

The important point is that you cannot go in the reverse direction: every root of $5-y^2=1$ is also a root of $5-(t-y^2)y^2=1$, but not the other way around, because there's no way to derive the former equation from the latter (unless you split into cases).

Answer 2

Let me correct a misconception of yours..?. It need not be nessessary that a quadratic eq should have only 2 roots... If u consider a quadratic eq in identity then it possesses not just 3 or 90... It possesses infinite number of roots I suppose u know what is a quadratic eq in identity.. Hope this helps..

Answer 3

By multiplying the same equation to your equation, you make a higher order equation but with unwanted roots. It is like saying $x=a$ so $(x-a)(x-a)=0*0\to x=+a,-a$

Let's make the problem simple by this example:

We can certainly say if $x=1$, so $x+1=2$ is also correct. They are both the same and not independent equations.

However, by multiplying them you make a higher order equation like below:

$(x)(x+1)=1*2=2 \to x^2+x-2=0$

The roots of this equation are $1$ and $-2$. But we know that $-2$ is not really "the root" and is an unwanted root.

The real reason is that you should not multiply the roots of a system only due to the fact that they are correct or equal to 1 or so. Because it is like saying $0*x=0*y$. Continueing our problem:

$x^2+x-2=0 \to (x-1)*(x-2)=0 \to$ $($any number$)*(x-1)=0$ [because $x-1=0$]