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September 23rd, 2017, 01:31 AM   #1
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Lightbulb Order of Operations comprehension question

Hello, this is my first post here, so please go easy on me! Thank you for your time.

As far as I understand:

if $\displaystyle x = a + b$
then $\displaystyle x^{2} = (a + b)^{2}$
not $\displaystyle x^{2} = a^{2} + b^{2}$
which I can easily check by asking does $\displaystyle 5^{2} = 3^{2} + 2^{2}$? No.

My question is does anyone have a way of explaining this property so that I might better understand *why* this property exists, beyond simply memorising it as a property of algebra? The reason I ask is because sometimes I make this mistake in algebraic manipulations; a specific example would be from physics:

since $\displaystyle v = u + at$
$\displaystyle v^{2} = (u + at)^{2} = u^{2} + 2at + a^{2}t^{2}$

However, if I am not careful, in practise I have previously resolved this as:
$\displaystyle v^{2} = u^{2} + a^{2}t^{2}$, since in my mind I am "just squaring both sides."

So my request is if anyone could give me an understanding to help cement this algebraic rule in my mind in a deeper way, so I don't make this mistake again.

I hope that's clear. Thanks in advance for any insight.

Sam
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October 1st, 2017, 01:15 AM   #2
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$(a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) = a*a + ab + b*a + b*b = a^2 + 2ab + b^2$
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October 1st, 2017, 01:20 AM   #3
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This mistake has a name of its own! See here. https://en.m.wikipedia.org/wiki/Freshman%27s_dream
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October 1st, 2017, 03:12 AM   #4
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Quote:
Originally Posted by sdparsons View Post
since $\displaystyle v = u + at$
$\displaystyle v^{2} = (u + at)^{2} = u^{2} + 2at + a^{2}t^{2}$
Should be: = u^2 + 2uat + a^2t^2

Code:
 2 + 3
 2 + 3
-------
4 + 6       : start with bottom left "2"
      6 + 9 : do bottom "3" next
-----------
4 +12 + 9 = 25
There's really 2 multiplications:
2 * (2 + 3), then 3 * (2 + 3)
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