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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 satisfying the system of equations, what is intuition behind the error in method 2?

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    $\begingroup$ Edited to correct MathJax, please refer to corrections and adapt in the future, you have just missed something quite tiny to be fair. On your question : If $a$ is a root of $p(x)$ and $q(x)$ then it's a root of $(p-q)(x)$, $(p+q)(x)$ and so on. But these polynomials could have other roots, which have nothing to do with $p$ or $q$. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Sep 1, 2021 at 16:03
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    $\begingroup$ Both methods are one way implications which are not reversible. Method 1, for example, means that if a common root exists, then it must be $x=-2$, but it does not prove that $x=-2$ is in fact a root. Try to apply method 1 to the system $x+3=0, 2x+1=0$ for example. $\endgroup$
    – dxiv
    Sep 1, 2021 at 16:12
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    $\begingroup$ Both methods are fine but to get an equivalence you need to keep one of the initial equations or another composition of these (keeping the difference and the sum is fine since you could recompose both initial equations!) $\endgroup$
    – Raymond Manzoni
    Sep 1, 2021 at 16:14
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    $\begingroup$ @DheerajGujrathi When you perform $p(x)-q(x)$, it is true that IF $a$ is a common root of $p(x)$ and $q(x)$, then it is also a root of $(p-q)(x)$, and of $(p+q)(x)$ , and of $(2p+3q)(x)$ and so on. But the opposite is not true : if I now take any root of $(p-q)(x)$, it is not necessary that it must have been a common root of $p(x)$ and $q(x)$. What is the reason? Well, think about numbers : $1$ is not a multiple of $3$ and $2$ is not a multiple of $3$, but $1+2 = 3$ is a multiple of $3$. So $1+2$, or $1-2$ , can only share some multiplicative property that both $1$ and $2$ share commonly $\endgroup$
    – Sarvesh Ravichandran Iyer
    Sep 1, 2021 at 16:27
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    $\begingroup$ The reason why this holds for numbers is because if $a$ and $b$ are two numbers, then all we can say about the divisors of $a+b$ for sure, is that IF $a,b$ share a divisor then this number is also a divisor of $a+b$. BUT we can't discuss factors of $a+b$ that don't come from either $a$ or $b$ ( to do this, one will have to discuss remainders, which is out of the present context). $\endgroup$
    – Sarvesh Ravichandran Iyer
    Sep 1, 2021 at 16:29

5 Answers 5

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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?

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  • $\begingroup$ ,yes,I can,thanks,but one more question,what if instead of 4x it was written ax,and we were told to find a such that it both polynomials share a common factor?we can't factor them out? $\endgroup$
    – Dheeraj Gujrathi
    Sep 1, 2021 at 16:39
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    $\begingroup$ If you had an unknown parameter $a$, you could solve the corresponding quadratic equation and compare the roots to those related to the second equation. In this way, you could find its value. $\endgroup$
    – user0102
    Sep 1, 2021 at 16:40
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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.

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The reason, that method 2 can lead to an extraneous solution, is that you replaced two equations by a single equation and in that you lost some information.

You had two equations: \begin{align*} x^2+4x+4&=0\\\\ x^2+5x+6&=0 \end{align*}

You replaced them by their sum: $$2x^2+9x+10=0.\tag{1}$$

But from the equation $(1)$ you cannot deduce the original two equations.

For example, if you used both the sum and one of the original equations \begin{align*} x^2+4x+4&=0\\\\ 2x^2+9x+10&=0 \end{align*} you would get an equivalent system of equations - since from these two equations you can derive the original equations. (By subtracting the first one from the second one.)

Naturally, simply replacing system of two equations can lead to an extraneous solution two. It wasn't the case here - but it can happen for various other systems of equations.

Still, I would not call this an error - one sholud keep in mind that if we're change an equation (or a system) to a new one which is not equivalent to the original one (but is consequence of the original one) there might be more solutions. So one should check (whenever using non-equivalent transformations) whether there aren't extraneous solutions.

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  • $\begingroup$ are you suggesting that those two equation represent specific necessity that is missing in third combined form,example a+b=c,a+2b=3c,these two equations imply two different necessities to be completed,but thier addition 2a+3b=4c does not state anything about the conditions on a+b and a+2b,the second equation might suggest a+2b=c and a+b=3c ,have I got it correct? $\endgroup$
    – Dheeraj Gujrathi
    Sep 10 at 5:52
  • $\begingroup$ @DheerajGujrathi This is exactly the idea I was trying to get across. Every solution of the system a+b=c, a+2b=3c is a solution 2a+3b=4c but not vice-versa. But if you compare the systems a+b=c, a+2b=3c and 2a+3b=4c, b=2c; these systems are equivalent and they have exactly the same solutions. I would suggest that if we want to discuss this further, we can try to continue in this chatroom - so that we do not create a long comment thread here. $\endgroup$
    – Martin Sleziak
    Sep 10 at 6:27
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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).

enter image description here

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.

enter image description here

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).]

enter image description here

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.

enter image description here

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$x^2+4x+4=0 \implies (x+2)^2=0$ 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 $b$

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