riosonicocyo

2022-03-16

Can the quadratic formula be explained intuitively?

varkdrafx9w

Step 1
The development of the quadratic formula is based on solving the quadratic equation in the form
$a{x}^{2}+bx+c=0$
By completing the square, we have
$a{x}^{2}+bx+c=0$
$a{x}^{2}+bx=-c$
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
${x}^{2}+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^{2}={\left(\frac{b}{2a}\right)}^{2}-\frac{c}{a}$
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$
$x=-\frac{b}{2a}±\frac{\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

elementalfoxwqe

Step 1
The equation
$a{x}^{2}+bx+c=0$
can be normalized to a simpler form by using a linear change of variable such as
$x=pt+q.$
Plugging in the equation, we get
$a{p}^{2}{t}^{2}+\left(2apq+bp\right)t+a{q}^{2}+bq+c=0.$
Now (WLOG $a>0$) we are free to set
$\left\{\begin{array}{l}a{p}^{2}=1,\\ 2apq+bp=0\end{array}$
and the equation simplifies to
${t}^{2}-d=0$
for some constant d.
Obviously the solutions are
$t=±\sqrt{d}$
and are real for $d>0$.
Hence the solutions in x are a linear function of $\mid ±\sqrt{d}$. Solving for the parameters p,q,d, you obtain the classical formulas.

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