Why is it sometimes possible to solve multi variable single equation?

Question

Assume you have an equation:

$\ 5x^2+4x+6=ax^2+bx+c $

Now theoretically, this is an equation of 4 variables, and it should not be solvable, but it is very apparent that a,b,c equals 5,4,6, in that order. Something similar happens with complex numbers:

$\ 5 + 4i = a + bi $

Here solution for a,b is 5,4, again it is very apparent.

My theory is that a,b or c, in these are something like incompatible number types. Therefore you can construct a system of equations from one single equation. The complex number equation is a good example, you can easily separate the complex and the non-complex part, as they virtually cannot influence the other.

But in the first equation, it's all in the real plane, there is no complex stuff going on and the x squared can influence the x

You can even do this:

$\ c = 5x^2 + 4x + 6 - ax^2 - bx $

Which would suggest c is dependent on the value of both a and b, and you get a similar result defining a or b, potentially pointing to an infinite number of solutions.

So why do these have only one solution? Assuming my theory is somewhat correct, what do mathematicians call these "incompatible numbers" properly?

Answer 1

The uniqueness in your examples is a consequence of some implicit assumptions in each case.

For the first, the unspoken assumption in the equation $$ 5x^2+4x+6=ax^2+bx+c $$ is that it is true for every value of $x$. That really makes it infinitely many equations. You need just three of them to determine $a$, $b$ and $c$. For example, setting $x$ to be $0$, $1$ and $-1$ in turn leads to $$ 6 = c. $$ $$ 5 + 4 + 6 = a + b + c $$ and $$ 5 -4 +6 = a - b + c . $$ It's no surprise that these determine $a$, $b$ and $c$.

In the second case, a complex number is usually defined as a pair of real numbers, the real and imaginary parts. It's no surprise then that two complex numbers are equal just when their real and imaginary parts match. This is just the assertion that if you knoa a point in the plane you know both its $x$ and $y$ coordinates.

Answer 2

Your question touches on a very important concept in linear algebra, that of linear independence. Some set of vectors $v_1,v_2,\dots,v_n$ are linearly independent if the only solution to $a_1v_1+a_2v_2+\dots+a_nv_n=0$ is when $a_1=a_2=\dots=a_n=0$. For example, $x^2$, $x$, and $1$ are linearly independent because if $f(x)=ax^2+bx+c$ and for any $x$, $f(x)=0$ (ie. $f$ is the $0$ function) then $a=b=c=0$. Now, a very important concept in linear algebra is that if we have some vector $v=a_1v_1+a_2v_2+\dots+a_nv_n$ there is only one possible choice of $a_1,a_2,\dots,a_n$. That means that if $ax^2+bx+c=dx^2+ex+f$, $a=d$, $b=e$, and $c=f$. In addition, note that $1$ and $i$ are linearly independent over the reals because the only real numbers $a,b$ where $a(1)+b(i)=0$ are $a=b=0$, so this property holds again. If you are interested in this topic, I recomend you research linear independence and linear algebra.

Answer 3

You are right when you say that you need more equations to find 4 unknowns. The issue is that you consider the equation $5x^2+4x+6=ax^2+bx+c$ to be verified for every possible value of $x$. Indeed, the solution $x=1,a=15,b=0,c=0$ works even though $a,b$ and $c$ are different from your values.