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Is there an easy way to square this equation?
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Is there an easy way to square this equation?
(ax^2+bx+c)? I always have to write it out and do it manually, is there a specific formula or something?
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>>7857713
why dont you compute it with a, b, c and see what comes out?
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>>7857717
With what? If it's the GC, i'm retarded with it, I do most of my shit with pen and paper, i use GC for Systems of Linear Equation and Basically observing graphs.
>>
[eqn](ax^2+bx+c)^2 = a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2[/eqn]
>>
>>7857726
what? with pen and paper bro. just expand it.

like (a+b)^2 = a(a+b) + b(a+b) = aa + ab + ba + bb = aa + 2ab + bb
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>>7857713
(ax^2 + bx + c)^2 = ( (ax^2+bx) + c ) ^2
The extra parenthesis is so you use the square binomial formula.

(x+y)^2 = x^2 + 2xy + y^2
where
x= ax^2+bx
y = c
>>
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>>7857713
>>
>>7857713
Sleep thight pupper
Sleep thight kitter
>>
>>7857713

I was doing similar hs-tier shit for mid-sized polynomials some months ago OP. You would be well-served to look up the version of the binomial theorem which (may) apply here.

Hint: the terms in your thing can be thought of as their own variables, which can then be plugged into whatever version of the above (often involving "x" and "y")

Also a nitpick, this isn't pure pedantry, it actually does matter: your original thing is not a equation (what is it supposed to be equal to? zero? I don't see that), but instead an /expression/. Squaring an expression is what we're doing.
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>>7857713
Use Vieta's method or how the fuck's that called
>>
>>7857731
>>
>>7857747
:D
>>
>ctrl+f
>pascals triangle
>0 results

i'm kinda mad about that /sci/.
Thread replies: 13
Thread images: 3
Thread DB ID: 513604



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