Why is any number squared equal to the the number before it times the number after it plus one?
Basically:
x(x+2)+1=(x+1)^2
Does this work for all numbers??
In: Mathematics
Expand it:
(X + 1)^2 = (x + 1)(x + 1) = x^2 + 2x + 1
Can you turn that into x(x + 2)+ 1 ?
These are both just different ways of writing x^2 + 2x + 1 (expand the brackets on both sides), so yeah, they’re always equal.
A clearer way to write it is:
> Basically:
> **(x-1)(x+1)** + 1 = x^^2
Since (x-1)(x+1) = x^^2 – 1
this can be simplified to:
> Basically:
> **x^^2 – 1** + 1 = x^^2
(a + b)^2
(a+b)(a+b)
a^2 + ab + ab + b^2 (FOIL)
a^2 + 2ab + b^2
This works for all a and b
You just have the case of b=1 and you factored out an a
a(a+2b) + b^2
Adding to the other answers, if you multiply the number two before and two after, the difference is 4. If you multiply the number three before and three after, the difference is 9. See the pattern?
Arrange a bunch of, say, coins in a square grid. If there are *n* rows and *n* columns of coins, there will be *n*^2 coins in total. Now remove one row of *n* coins and use these to add an extra column. But, because you now have one fewer row, you only need *n*−1 coins for the new column. You have one coin left over. This demonstration works for any whole number *n* from 2 up.
Just to pile on, multiplication is just compact notation for addition. 5*5 is a compact notation of 5+5+5+5+5. Now imagine it’s a big number and you don’t have a calculator, if you remove a 5 what happens? It goes down by 5. What happens if you change all the remaining 5s into 6s? It goes up by the number of numbers. Put it together: if you take a square, lower one factor and raise the new opposite the result is a subtraction of the original factor and addition of the exactly once lowered factor. So 5^2 minus 5 plus (5-1) is 24 or (5^2 )-1.
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