Deceptively Simple System of Equations

I’m currently taking Differential Equations, and I have never felt more mathy. That doesn’t mean I don’t occasionally get stuck on simple algebra. Consider these equations from a homework problem:

$$ 2a_{2} + 3a_{1}x + 4a_{0} = 0 $$

$$ a_{2} + a_{1} + a_{0} = 1 $$

$$ \text{where } a_{0}, a_{1}, a_{2} \text{ are constants} $$

System of equations are introduced in high school, so I’ve solved tons of these problems before using the elimination technique (I don’t know any Linear Algebra yet).

I had given up on this problem until my professor (and a friend) showed me a couple of tricks.

$$ 2a_{2} + 3a_{1}x + 4a_{0} = 3a_{1}x + (2a_{2} + 4a_{0}) = 0 $$

It’s important to note that this function is a 1st degree polynomial. That means:

$$ 3a_{1}x + (2a_{2} + 4a_{0}) = 0x + 0 $$

The coefficients in a polynomial are just constants. Consider (3a*{1}x). The coefficient (3a*{1}) is just some number (like 5 or 100), and can never be something like (1/x) and cancel out (x). Similarly, combining (2a*{2} + 4a*{0}) will always result in a number since two constants can’t result in an (x) or (x^2) for example.

Therefore, the only way to get (0x) is if (3a*{1} = 0). And the only way to get (0) (referring to the last term of the equation on the right) is if (2a*{2} + 4a_{0} = 0).

The pattern should be apparent now. For every term of degree (n), we equate the coefficient of the left side to the coefficient on right side of the equation:

$$ 3a_{1} = 0 $$

Where (3a*{1}) is the coefficient of (3a*{1}x) and (0) is the coefficient of (0x).

$$ 2a_{2} + 4a_{0} = 0 $$

Where (2a*{2} + 4a*{0}) and (0) are the constants being added to both equations.

The rest of this problem is familiar territory. By simplifying the equations above, we get:

$$ a_{1} = 0 $$

$$ a_{2} = -2a_{0} $$

If we plug these values into the original equation ((a*{2} + a*{1} + a_{0} = 1)), we get:

$$ -2a_{0} + 0 + a_{0} = 1 $$

And we finally get the answer!

$$ a_{0} = -1 $$

$$ a_{1} = 0 $$

$$ a_{2} = 2 $$