<s>[INST] 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?
 [/INST] 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.

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