Why does Solving system of quadratic equations gives extra roots? Consider these system of Equations
\begin{align*}
\begin{cases}
x^2+4x+4=0\\\\
x^2+5x+6=0
\end{cases}
\end{align*}
For solving them
We have
Method 1-
Subtract both equations
So
$-x-2=0$
Hence,
$x=-2$
Method-2
Add both equations
$2x^2+9x+10=0$
After applying quadratic formula,we get
$x=-2$ or $x=-5/2$. But only $x=-2$ satisfies the system of equation.
Why is the $-5/2$ not satisying the system of equations,what is intuition behind the error in method 2?

We could also describe the difference in your two approaches in this way.  You began with the system $ \ p(x) \ = \ x^2   + 4x + 4 \ = \ 0 \ , \ q(x) \ = \ x^2 + 5x + 6  \ = \ 0 \ \ . $  When you subtract  one equation from the other, you have $ \ p(x) - q(x) \ = \ 0 \ \ , $ which is equivalent to the equation we would set up for finding intersections of the curves represented by these functions, $ \ p(x) \ = \ q(x) \ \ . $  You found the single solution $ \ x \ = \ -2 \ \   $ from $ \ -x - 2 \ = \ 0 \ \ , $ which is correct.  It so happens for this system that this also locates a common factor of $ \ p(x) \ $ and $ \ q(x) \ $ because both functions are equal to zero at $ \ x \ = \ -2 \ \ , $ but this method would be correct to use in any case (as we'll show shortly for a different system).

When you add the two equations, as you did in your second calculation, you are now solving $ \ p(x) + q(x) \ = \ 0 \ \rightarrow \ p(x) \ = \ -q(x) \ \ , $ which is no longer the original problem.  Here, it  really is  only because both $ \ p(x) \ $ and $ \ q(x) \ $ are equal to zero at $ \ x \ = \ -2 \ $ that this appears as a solution to $ \ 2x^2 + 9x + 10 \ = \ (2x + 5 )·(x + 2) \ = \ 0 \ \ , $ since $ \ (x + 2)·(x + 2) \ + \ (x + 2)·(x + 3) $
$ = \ (x + 2)·[ \ (x + 2) + (x + 3) \ ] \ \ . $  Other systems may not produce any solutions at all by adding the equations.

If we take, for example, the system $ \ p(x) \ = \ x^2 - 5x + 6  \ = \ ( x - 2)·(x - 3) \ = \ 0 \ , $
$ q(x) \ = \ x^2   + 7x + 10 \ = \ (x + 2)·(x + 5) \ =  0 \ \ , $ the two polynomials have no factors in common, but surely must intersect since they both "open upward".  We find $ \ p(x) \ = \ q(x) $ $ \Rightarrow \ p(x) - q(x) \ = \ -12x - 4 \ = \ 0 \ \Rightarrow \ x \ = \ -\frac13 \ \ , $ a solution which is not suggested immediately by any of the polynomial factors.  [Indeed, we could have chosen two "upward-opening" parabolas represented by polynomials which cannot be factored using real numbers and would still be able to find the intersection(s).]

On the other hand, $ \ p(x) + q(x) \ = \ 2x^2 + 2x + 16 \ \ $ has no real zeroes (or $ \ p(x) \ = \ -q(x) \ $ has no real-number solutions); we see that the function curves do not intersect.  So adding the equations together in this system provides no information about the solutions of the original system of equations.

