riosonicocyo

2022-03-16

Can the quadratic formula be explained intuitively?

varkdrafx9w

Beginner2022-03-17Added 4 answers

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

$a{x}^{2}+bx+c=0$

$a{x}^{2}+bx=-c$

$x}^{2}+\frac{b}{a}x=-\frac{c}{a$

$x}^{2}+\frac{b}{a}x{+{\left(\frac{b}{2a}\right)}^{2}}={{\left(\frac{b}{2a}\right)}^{2}}-\frac{c}{a$

$(x+\frac{b}{2a})}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}$

$x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$

$x=-\frac{b}{2a}\pm \frac{\sqrt{{b}^{2}-4ac}}{2a}$

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

The development of the quadratic formula is based on solving the quadratic equation in the form

By completing the square, we have

elementalfoxwqe

Beginner2022-03-18Added 3 answers

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

$a{p}^{2}{t}^{2}+(2apq+bp)t+a{q}^{2}+bq+c=0.$

Now (WLOG$a>0$ ) we are free to set

$\{\begin{array}{l}a{p}^{2}=1,\\ 2apq+bp=0\end{array}$

and the equation simplifies to

${t}^{2}-d=0$

for some constant d.

Obviously the solutions are

$t=\pm \sqrt{d}$

and are real for$d>0$ .

Hence the solutions in x are a linear function of$\mid \pm \sqrt{d}$ . Solving for the parameters p,q,d, you obtain the classical formulas.

The equation

can be normalized to a simpler form by using a linear change of variable such as

Plugging in the equation, we get

Now (WLOG

and the equation simplifies to

for some constant d.

Obviously the solutions are

and are real for

Hence the solutions in x are a linear function of

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