/sci/ - Science & Math » Thread #12813201

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Anonymous Wed 10 Mar 00:43:56 2021 No.12813201 View Reply Original Report

Quoted By: >>12813205 >>12813284

>he posts on /sci/
>he thinks hes smart
>passed his college math classes
>he can't derive the quadratic formula
this isn't you right /sci/? I hope none of you are dumber than a high school freshman.

Anonymous

Anonymous Wed 10 Mar 2021 00:44:52 No.12813205 Report

Quoted By: >>12813219

>>12813201
whom are you talking to, schizo?

Anonymous

Anonymous Wed 10 Mar 2021 00:47:12 No.12813219 Report

Quoted By:

>>12813205
>t. can't derive the quadratic formula

Anonymous

Anonymous Wed 10 Mar 2021 01:01:00 No.12813277 Report

Quoted By:

Deriving the quadratic formula is trivial, faggot. Observe:
$ax^2 + bx + c = 0$
can be recast as
$k(x - n)^2 + m= 0$
From which we trivially find
$x = n /pm /sqrt{/frac{-m}{k}}$ .
Done.

Anonymous

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Anonymous Wed 10 Mar 2021 01:02:07 No.12813284 Report

Quoted By:

>>12813201
Remember, no credit if it isn't stable

Anonymous

Anonymous Wed 10 Mar 2021 01:15:21 No.12813348 Report

Quoted By:

An unorthodox derivation, from someone who doesn't remember that faggy "completing the square":

We begin with the standard form quadratic,
$ax^2 + bx + c = 0 \rightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
Next, we write it in vertex form:
$(x - h)^2 + k = 0$
recognizing that
$\begin{align}<br />h &= \frac{b}{2a} \\<br />k &= \frac{c}{a}-\frac{b^2}{4a^2}<br />\end{align}<br />$
Finally, we solve
$x = h \pm i\sqrt{k}$
Q.E.D.

You can of course, write this in standard fag form, but that's your problem.

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