Bored in math class again, so I made this

[Blahness]

If B - A = 1,
Then
A^2 + A + B = B^2.

Good for figuring out an exponent next to one you know.
Example:
You know 50^2 is 2500, but need 49^2.

B^2(2500) - B(50) - A(49) = B^2(2401).

Make sense?
Useful, useless, w/e?

 

[TD]

Make sense?

Sure:

a^2 + a + b = b^2 \Leftrightarrow a^2 - b^2 = - \left( {a + b} \right) \Leftrightarrow \left( {a - b} \right)\left( {a + b} \right) = - \left( {a + b} \right) \Leftrightarrow \left( {a - b} \right) = - 1 \Leftrightarrow b - a = 1

Useful, useless, w/e?

I suppose so, but I doubt it's "new"

 

[Blahness]

Probably not, but I need to know that the crap I think up in math class isn't junk! ^_^
Ty ^_^

 

[Galileo]

Knowing and applying 'tricks' like these is usually how I am able to multiply big numbersin my head. Note: Big means 2 digits.
For example, using (a+b)(a-b)=a^2-b^2, calculating 47 times 53 is easy:
47 \cdot 53=(50+3)(50-3)=50^2-3^2=2500-9=2491

 

[Kamataat]

To find roots near 50, use (50+/-x)^2 = 2500 +/- 100x + x^2. In other words, to find 47^2 just subtract 3 from 25 to get 22 and square 3 to get 09, so 47^2=2209.

Read it from one of Feynman's books.

- Kamataat

 

[masudr]

Unfortunately, these sorts of tricks alone will not get you very far.

 

[Galileo]

They always work like a charm for me. Then again, I really suck at mental calculations without 'tricks'.

To square a number a ending with a five quickly:
Write a = 10b+5. (division by 10 with remainder 5). b is simply the number you get after dropping the 5 mentally.
a^2=(10b+5)^2=100b^2+100b+25=100b(b+1)+25
So you simply take b, multiply with the next integer and glue 25 at the end.

25^2: 2 times 3 is 6. 'add' 25 to get 625
85^2: 8 times 9 is 72. 'add' 25 to get 7225
etc.

 

[quasar987]

To find roots near 50, use (50+/-x)^2 = 2500 +/- 100x + x^2. In other words, to find 47^2 just subtract 3 from 25 to get 22 and square 3 to get 09, so 47^2=2209.

Read it from one of Feynman's books.

- Kamataat

I think this trick given to Feynman by Hans Bethe while they were at Los Alamos!

 

[MathematicalPhysicist]

They always work like a charm for me. Then again, I really suck at mental calculations without 'tricks'.

etc.

mental calculations as you coined it really depend on your memory.

 

[Cosmo16]

Another one I have found is this.

5^2= 25 0=n
15^2= 225 2=n
25^2= 625 6=n
35^2=1225 12=n
45^2=2025 20=n
55^2=3025 30=n

How would I express that algebracly?

 

[shmoe]

Try expanding (m*10+5)^2

 

[Kamataat]

I think this trick given to Feynman by Hans Bethe while they were at Los Alamos!

Yes, it was indeed!

- Kamataat

 

[Robokapp]

My achievements in math: a^2=(a+1)*(a-1)+1 for {a>N/a>0}

in other words 49^2=50*48

so 49^2=2400

 

[roger]

My achievements in math: a^2=(a+1)*(a-1)+1 for {a>N/a>0}
in other words 49^2=50*48
so 49^2=2400

You forgot to add 1.

 

[Werg22]

My achievements in math: a^2=(a+1)*(a-1)+1 for {a>N/a>0}
in other words 49^2=50*48
so 49^2=2400

A power has the same factors as its rational roots.

 

[Blahness]

49^2 = 2401.

Anyway, if it's a^2=(a+1)*(a-1)+1
That becomes
49^2=(50)*(48)+1
reduces to
2401 = 2401.

Just clearing that up. ^_^

Much more simplistic version of my equation, nice Robo! ^_^''