I wonder if this is a valid method to derive the quadratic formula.
It is not an optimal method at all, but I am intrigued if it is a genuine way or has any gap.
The method starts from the equation whose coefficients are such that it is not necessary to use the quadratic formula to solve it,
$$(a^2 + b^2)x^2+ 2bcx+c^2 =0$$ becouse we can easily solve for x, having at the right side a quadratic form: $$a^2x^2 = -b^2x^2 – 2bcx – c^2 \implies a^2x^2 = -(bx + c)^2 $$
Obtaining the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is trivial and the coefficient $a$ is squared so that $(a^2 + b^2)$ is only zero when $a$ or $b$ are zero. Of course $a,b,c \in \mathbb{R}$
First we find the roots of our special equation: $a^2x^2 = -(bx + c)^2 $
Square rooting both sides: $ax = \pm i(bx + c) $ where $ i=\sqrt{-1}$
Solving x we obtain: $ x = \dfrac{\pm ic}{a \mp ib}$
Now we are ready to derive the general quadratic formula. Note that the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is just a quadratic equation $a'x^2 + b'x +c'$ so we can rearrange the coefficients:
$$a'=(a^2 + b^2)$$
$$b'=2bc$$
$$c'= c^2$$
Let's solve for $a$, $b$ and $c$
$$c'= c^2 \implies c = \pm \sqrt{c'} $$
$$b'=2bc \implies b= \frac{b'}{\pm 2\sqrt{c'}} $$
$$a'=(a^2 + b^2) \implies a = \pm \sqrt{a' – \frac{b'^2}{4c'} } $$
And now we substitute $a$, $b$ and $c$ in the above result $ x = \dfrac{\pm ic}{a \mp ib}$ and we obtain:
$$ x = \frac{\pm i \sqrt{c'}}{\pm \sqrt{a' – \frac{b'^2}{4c'} – \mp i \frac{b'}{2\sqrt{c'}} }}$$
Multiplying top and bottom by $\pm \sqrt{c'}$
$$ x = \frac{\pm i c'}{\pm \sqrt{c'} \sqrt{a' – \frac{b'^2}{4c'} – \mp i \frac{b'}{2} }}$$
Taking $4c'$ out of the square root and simplifying
$$ x = \frac{\pm 2 i c'}{\pm \sqrt{4a'c' – b'^2} \mp i b'}$$
Multiplying up and down by the conjugate of the denominator:
$$ x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' – b'^2} \mp i b')}{(\pm \sqrt{4a'c' – b'^2} \mp i b')(\pm \sqrt{4a'c' – b'^2} \pm i b')} \implies$$
$$ \implies x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' – b'^2} \mp i b')}{4a'c'}$$
Note that $ i \sqrt{4a'c' – b'^2} = \sqrt{b'^2 – 4a'c'}$ and $\pm i * \pm ib' = -b'$ So simplifying:
$$ x = \frac{-b'}{2a'} \pm \frac{\pm \sqrt{b'^2 – 4a'c'} }{2a'}$$
And, we've derived the formula.
Best Answer
[Edit] The following assumes a proper quadratic with non-zero roots, meaning that both the leading coefficient and the constant term are non-$0\,$ i.e. $\,a'c' \ne 0\,$ with OP's notations. These are the same assumptions (implicitly) made in the original post.
The method is generally correct, after fixing an oversight and a couple of typos. It is, however, essentially equivalent to the traditional method of "completing the square", only applied to the quadratic $\,c'x^2+b'x+a'\,$ obtained by the substitution $\,x \to 1/x\,$.
It is important to note here that the $\,\pm\,$ signs for $\,b,c\,$ correspond, while the $\,\pm\,$ sign for $\,a\,$ is independent and uncorrelated to the other two.
This should rather be:
$$ x = \frac{\pm i \sqrt{c'}}{\sqrt{a' - \dfrac{b'^2}{4c'}} \;\mp\; i \dfrac{b'}{2\sqrt{c'}}}$$
Why:
this fixes the typo which extended the square root over the entire denominator;
it removes the extraneous $\,\pm\,$ sign before the radical, which was not necessary since changing the sign of the radical is equivalent to flipping the signs of the other two terms $\;-\;$ which also removes the ambiguity of the posted formula appearing to define $\,2\cdot 2 = 4\,$ values of $\,x\,$, while in reality it's only $\,2\,$ of them.
About the similarity with the traditional method of "completing the square", OP's approach is essentially equivalent to the following:
substitute $\,y=1/x\,$ in $\,a'x^2+b'x+c'=0\,$ and derive $\,c'y^2+b'y+a'=0\,$;
write the quadratic as $\,c''^{\,2} y^2 + 2c''b''y +b''^{\,2} + a''\,$ for suitably chosen $\,c'',b'',a''\,$;
solve $\,(c''y+b'')^2+a''=0\,$ for $\,y\,$, then calculate $\,x = 1/y\,$.