"Can the quadratic formula be explained intuitively?" solved and explained

riosonicocyo

Answered question

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} + b x + c = 0

By completing the square, we have

a x^{2} + b x + c = 0

a x^{2} + b x = - c

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

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

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

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

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

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

elementalfoxwqe

Beginner2022-03-18Added 3 answers

Step 1
The equation

a x^{2} + b x + c = 0

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

x = p t + q .

Plugging in the equation, we get

a p^{2} t^{2} + (2 a p q + b p) t + a q^{2} + b q + c = 0 .

Now (WLOG

a > 0

) we are free to set

{\begin{cases} a p^{2} = 1, \\ 2 a p q + b p = 0 \end{cases}

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

∣ \pm \sqrt{d}

. Solving for the parameters p,q,d, you obtain the classical formulas.

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