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I have a system of two quadratic equations $$ \left\{ \begin{array}{c} 2x^2+x-1=0 \\ 2x^2+5x+2=0 \end{array} \right. $$

I tried to solve it the following way: $$ 2x^2=-5x-2$$ substituting in the first equation $$ -5x-2+x-1=0$$ $$ -4x-3=0$$ $$ x=-3/4$$ However, this result just makes the two expressions equal to each other, but not equal to zero. Under another question someone suggested that equating the two expressions is a way to solve a system of quadratic equations. However that is precisely my problem. Where and why do I lose the information about zero? Is there a way to solve this that would tell me there is actually no solution? (Besides just solving them separately).

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  • $\begingroup$ I think the issue is that in the calculations following the substitution, there is an assumption that there EXISTS a common solution (i.e. among the values of $x$ that make the first equation true there will be at least one value that also makes the second equation true), and in this case that is a false assumption, and so by formal logic there is nothing incorrect about the logical steps no matter what you wind up concluding (i.e. "false" implies "anything" in logic). Anyway, that's my initial gut reaction, but I don't have time to flush this out better, so anyone else is welcome to continue. $\endgroup$ – Dave L. Renfro Dec 1 '18 at 16:56
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When you set the two equations equal to each other, you’re finding the $x$-coordinate where the graphs intersect, or the common point, not where $x$ gives an output of $0$ for both equations, hence the loss of information. This would apply regardless of the value of $c_1$ in

$$\left\{ \begin{array}{c} 2x^2+x-1=c_1 \\ 2x^2+5x+2=c_1 \end{array} \right.$$

Now, if you want to see where they intersect and give an output of $0$, you could use substitution for $x$ by solving for it in one of the equations.

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

$$x_1 = \frac{1}{2} \quad x_2 = -1$$

Now, plugging these values in the second equation, you get

$$2\bigg(\frac{1}{2}\bigg)^2+5\bigg(\frac{1}{2}\bigg)+2 = 5 \color{red}{\neq 0}$$

$$2(-1)^2+5(-1)+2 = -1 \color{red}{\neq 0}$$

Hence, there is no common root. You would notice the same if you choose any $c_1$, as mentioned. You would solve for the common point at $x = -\frac{3}{4}$, but plugging in that value wouldn’t yield $c_1$.

I’m not sure if this is the best explanation or if it precisely answers the question, but this is what I get from it.

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[Edit: This explains why there needn't be a solution. See my other answer for what went wrong in trying to find one.]

The problem you've got is that either equation can already be fully solved on its own. Each quadratic expression on the left defines a parabola, and the solutions to each equation are the points where its parabola intersects the $x$ axis.

So there would only be a common solution if both parabolas happened to intersect the $x$ axis in the same place.

As noted in another answer, trying the roots of one equation as values of $x$ in the other tells you whether this actually happens—and in this case it doesn't.

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Quite simple: you can apply the rational roots theorem and check $-1$ is a root of the first equation, so by Vieta's relations $\frac12$ is the other root. None of these numbers is a root of the second equation.

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    $\begingroup$ This seems to address the issue of how to find the other root(s), rather than (what was asked) what caused the other root to be lost, since it seems everything involved making legitimate mathematical steps. $\endgroup$ – Dave L. Renfro Dec 1 '18 at 16:52
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OK, now I see what's actually happened. Try replacing $0$ with $y$ in both equations.

What you've actually done is to eliminate $y$. Instead of $$-4x-3=0$$ you'd have $$y-4x-3=y$$ after your substitution, then subtract out the $y$ and find $$x=-\frac{3}{4}$$ as you did.

Next you'd substitute this in either of the original equations to get $$y=-\frac{5}{8}$$

So what you've effectively done is solve two equations in $x$ and $y$, but use $0$ as the name for $y$.

You've treated $0$ as a variable. The information about $0$ disappeared when you eliminated the variable to find $x$.

When you checked by putting $x$ into the original expressions, you got the information about $0$ back. And the information was: $$0=-\frac{5}{8}$$ which demonstrated the inconsistency of the system of equations.

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The system of equations is equivalent to: $$2x^2+x-1=2x^2+5x+2=0.$$ You solved the LHS equation and found $x=-\frac34$. And now you must make sure the same number suits the RHS equation. (Or you can solve the RHS equation and compare the roots). The solution of a system of equations is such a number that suits all equations.

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  • $\begingroup$ @timtfj. How many equations and unknowns did I write in line 2? Is it equivalent to the OP's system? What will be the conclusion if no number is found that suits all (in this case two) equations (of the system)? $\endgroup$ – farruhota Dec 2 '18 at 12:06
  • $\begingroup$ Normally, a system of $n$ equations is a set of $n$ unknowns. He's got two equations but only one unknown—meaning the equations are either dependent or inconsistent. Since the only $x$ satisfying the left equation turns the right one into $0=-\frac{5}{18}$, they're inconsistent. $\endgroup$ – timtfj Dec 2 '18 at 12:09
  • $\begingroup$ Yes, it's equivalent and is two equations (one for each equals sign) but only one unknown, $x$ $\endgroup$ – timtfj Dec 2 '18 at 12:12
  • $\begingroup$ (Comments are out of order because I couldn't edit my first comment and ended up deleting and replacing it while farruhota was replying to it. Apologies.) $\endgroup$ – timtfj Dec 2 '18 at 12:16
  • $\begingroup$ @timtfj, I don't understand your point: are you objecting my answer, confirming it or not understanding it? $\endgroup$ – farruhota Dec 2 '18 at 12:19

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