Information lost in solving system of quadratic equations

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

[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.

Answer 4

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.

Answer 5

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.