SOLUTION: Trying to help my friend with his algebra, and I'm stumped as to what they're going for on this one: Explain why: {{{(sqrt(n)-sqrt(n+1))(sqrt(n)-sqrt(n-1))=1}}} is true for {{{n

Algebra ->  Algebra  -> Square-cubic-other-roots -> SOLUTION: Trying to help my friend with his algebra, and I'm stumped as to what they're going for on this one: Explain why: {{{(sqrt(n)-sqrt(n+1))(sqrt(n)-sqrt(n-1))=1}}} is true for {{{n      Log On

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Question 226228: Trying to help my friend with his algebra, and I'm stumped as to what they're going for on this one:
Explain why:
%28sqrt%28n%29-sqrt%28n%2B1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1
is true for n%3E=1 two different ways.
a) Provide two numerical values.
b) Show that is it true in general.

Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
The equation should be %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1. Make sure that you have the correct equation.


a) Let's show that %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1 is true for the values n=1 and n=2


Let's show the equation is true for n=1:


%28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1 Start with the given equation.


%28sqrt%281%29%2Bsqrt%281-1%29%29%28sqrt%281%29-sqrt%281-1%29%29=1 Plug in n=1


%28sqrt%281%29%2Bsqrt%280%29%29%28sqrt%281%29-sqrt%280%29%29=1 Combine like terms.


%281%2B0%29%281-0%29=1 Take the square root of 1 to get 1. Take the square root of 0 to get 0.


%281%29%281%29=1 Combine like terms.


1=1 Multiply.


So %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1 is true when n=1


-----------------------



Let's show the equation is true for n=2:


%28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1 Start with the given equation.


%28sqrt%282%29%2Bsqrt%282-1%29%29%28sqrt%282%29-sqrt%282-1%29%29=1 Plug in n=1


%28sqrt%282%29%2Bsqrt%281%29%29%28sqrt%282%29-sqrt%281%29%29=1 Subtract


%28sqrt%282%29%2B1%29%28sqrt%282%29-1%29=1 Take the square root of 1 to get 1.


sqrt%282%29%2Asqrt%282%29-sqrt%282%29%2Bsqrt%282%29%2B1%28-1%29=1 FOIL


2-sqrt%282%29%2Bsqrt%282%29-1=1 Multiply


1=1 Combine like terms.


So %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29-sqrt%28n-1%29%29=1 is true when n=2


====================================================================

b)


Now let's show that %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29%2Bsqrt%28n-1%29%29=1 is true for any value of 'n' such that n%3E=1


%28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29%2Bsqrt%28n-1%29%29=1 Start with the given equation.


Let x=sqrt%28n%29 and y=sqrt%28n-1%29


%28x%2By%29%28x-y%29=1 Replace each sqrt%28n%29 with 'x' and replace each sqrt%28n-1%29 with 'y'


x%5E2-xy%2Bxy-y%5E2=1 FOIL


x%5E2-y%5E2=1 Combine like terms.


%28sqrt%28n%29%29%5E2-%28sqrt%28n-1%29%29%5E2=1 Plug in x=sqrt%28n%29 and y=sqrt%28n-1%29


n-%28n-1%29=1 Square each square root to eliminate them (note: because n%3E=1, this means that n-1%3E=0 which means that each radicand is positive. So we can avoid absolute values here).


n-n%2B1=1 Distribute


1=1 Combine like terms.


So this shows that %28sqrt%28n%29%2Bsqrt%28n-1%29%29%28sqrt%28n%29%2Bsqrt%28n-1%29%29=1 is true for any value of 'n' such that n%3E=1