I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.
Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)
We want to solve the equation
$$ax^2+bx+c=0,$$
where $a \ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.
Instead, multiply both sides by $4a$. We obtain the equivalent equation
$$4a^2x^2 +4abx+4ac=0.\tag{1}$$
Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
$$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
So our equation can be rewritten as
$$(2ax+b)^2 -b^2+4ac=0 \tag{2}$$
or equivalently
$$(2ax+b)^2=b^2-4ac. \tag{3}$$
Now it's all over. We find that
$$2ax+b=\pm\sqrt{b^2-4ac} \tag{4}$$
and therefore
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \tag{5}$$
No fractions until the very end!
Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.
Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.
This works in any base because $\log_b (a^c) = c \log_b(a)$
The practical reason for using base $10$ was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction. Calculators then provided a $\log_{10}$ function as a carry-over from tables and just called the button log
My 1953 elementary statistical tables have logarithms of numbers from $1$ to $10$ and anti-logarithms of numbers from $0$ to $1$. Since these are logarithms base $10$, I can easily deal with all numbers because $\log(a \times 10^n)=n +\log(a)$
Here I want $\dfrac{\log 7}{\log 2}$.
My tables tell me $\log 7\approx 0.8451_6$ (with the ${\,}_6$ helping interpolation) and $\log 2\approx 0.3010_{22}$. So that leaves me with trying to calculate $\dfrac{0.8451}{0.3010}$. I cannot be bothered to do long division, so instead I try to calculate $\text{antilog}\,({\log 0.8451 -\log 0.3010})$.
My tables tell me $\log 8.45 \approx 0.9269_5$ so I write $\log 0.8451 \approx \bar{1}.9269$ (with the $\bar{1}$ because I wanted $\log (8.451\times 10^{-1})= -1+\log 8.451$). Similarly it tells me $\log 3.01 \approx 0.4786_{14}$ so I write $\log 0.3010 \approx \bar{1}.4786$
I now calculate by hand $\bar{1}.9269 - \bar{1}.4786 = 0.4483$. My tables tell me $\text{antilog}\,0.448 \approx 2.805_7$ and I interpolate the final $3$ using the ${\,}_7$ to give a final approximate answer of $2.807$. Which is what you got with some clever silicon
Best Answer
If the bases are the same (and positive) then $a^x=a^y$ can be divided through by the non-zero number $a^y$ to obtain $a^{x-y}=1$ from which we can conclude that either $a=1$ (we assumed $a\gt 0$) or $x-y=0$.
[The function $a^x$ is monotonic for positive $a$ - it increases with increasing $x$ for $a\gt 1$, is constant for $a=1$ and decreases with increasing $x$ for $a\lt 1$ - this makes it injective for positive $a\neq 1$]
If the bases are not the same - well $4^2=16$ and $2^2=4$ are not equal, even though they have the same exponent. One can take logarithms to any convenient base, though, and in appropriate cases this will lead to a solution.