A linear equation is $$ ax + b = 0 ; \,\, \,\, a\neq 0 $$
A quadratic equation is $$ax^2 + bx + c = 0 ; \,\, a\neq 0 $$
And so on…
Why are all these equations written as $\dots = 0 $? Why do mathematicians do it this way?
[Math] Why are equations written by equating something to zero
algebra-precalculusnotationquadratics
Related Solutions
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.
It's the same thing. What Kline is doing is what's in general known as a depression, which is a variable substitution done for a polynomial of degree $n$, such that the resulting polynomial has no term of degree $x^{n-1}$.
One way to see how depression works is to consider the Vieta formulae, which in the case of the quadratic $ax^2+bx+c=a(x-x_1)(x-x_2)$ looks like this:
$$\begin{align*} x_1 x_2 &= \frac{c}{a}\\ x_1 + x_2 &= -\frac{b}{a} \end{align*}$$
From this, if the sum of the roots is $-\dfrac{b}{a}$, then the mean of the roots is $-\dfrac{b}{2a}$. Thus, the depression substitution
$$y=x+\frac{b}{2a}$$
can be geometrically interpreted as shifting the parabola corresponding to your quadratic $ax^2+bx+c$ such that it is "centered" about the origin, and the two roots are laid out symmetrically.
If we depress our original quadratic, we get
$$\require{cancel}\begin{align*} ax^2+bx+c&=a\left(y-\frac{b}{2a}\right)^2+b\left(y-\frac{b}{2a}\right)+c\\ &=a\left(y^2-\frac{b}{a}y+\frac{b^2}{4a^2}\right)+b\left(y-\frac{b}{2a}\right)+c\\ &=ay^2\cancel{-by}+\frac{b^2}{4a}\cancel{+by}-\frac{b^2}{2a}+c\\ &=ay^2+\frac{b^2}{4a}-\frac{2b^2}{4a}+c\\ &=ay^2-\frac{b^2}{4a}+c=ay^2-\frac{b^2-4ac}{4a} \end{align*}$$
and I'm sure you know how easy it is to solve the equation
$$ay^2-\frac{b^2-4ac}{4a}=0$$
Having solved for $y$, you undo the depression you did, which means you have to add the term $-\dfrac{b}{2a}$ to get the actual roots you want. That's where that part of the quadratic formula comes from.
In short, I would not say that Kline's method is the most expedient, but it at least looks to me that this slower method allows for more cogitation on what the steps are supposed to "mean", as opposed to a lazy plug-and-chug.
Best Answer
At least for quadratics, if you want to solve (for example) $x^2 + 5x +8 = 2$, it is much easier to subtract 2 from each side, and factor:
$x^2 +5x +8-2 =0$
$x^2 + 5x +6 =0$
$(x+2)(x+3)=0$
Here is the key: the only way for a product of numbers ($x+2$) and ($x+3$) to be equal to zero is for one to be zero. This is a property unique to zero, and explains (at least in part) why we often set equations equal to zero.