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?
 A: HINT
You can factor both polynomials according to your preferred method in order to obtain:
\begin{align*}
\begin{cases}
x^{2} + 4x + 4 = 0\\\\
x^{2} + 5x + 6 = 0
\end{cases} \Longleftrightarrow
\begin{cases}
(x+2)^{2} = 0\\\\
(x+2)(x+3) = 0
\end{cases}
\end{align*}
Can you take it from here?
A: To answer your title question:

Why does Solving system of quadratic equations gives extra roots?

This is because any quadratic equation can have at most two solutions, so a system of quadratic equations may at most two solutions in common between the two quadratic equations.
Specifically, you have two quadratic equations, and they share one solution, up to multiplicity, and so you have 3 extraneous roots.
If they have more than one solution in common, they'd necessarily be scalar multiples of one another.
A: 
 $x^2+4x+4=0 \implies (x+2)^2=0$ Similarly factorise the other polynomial and see which value of $x$ is common in them

For other part of question your method didn't gave the right answer because you must remember $(a-b)x,(a+b)x$ can have roots other than $a$ and $b$
A: 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.

