Solve this root

Question

Solve :  (sqrt(n+1))^2

Can someone help me solve this root? Please show work. Thanks.

timetrader · Answer

(\sqrt{n+1})^2 = |n+1|

I3igDmsu · Answer

Could someone show work please?

timetrader · Answer

sorry I didn't have any middle steps ... it kind of popped out of my head...

(\sqrt{n+1})^2 = n+1,  where n+1<0, has no solution, for real numbers (cuz  \sqrt{n} = n^{1/2},  where n is a positive real number... so  ((n+1)^{1/2})^2 = (n+1)^{{1/2*2}} =n+1 )
BUT
 (\sqrt{n+1})^2 = |n+1|  for complex numbers.

I guess you would need someone with degree in math to provide more insight to why it is like that, I'm not that good ))

I3igDmsu · Answer

I was looking to simplify the equation above, rather than solve. Can you use the FOIL method here? I guess what I am getting at is if you have an expression under a root and that expression is squared, then do you FOIL out the two expressions? How do you FOIL with roots?

timetrader · Answer

You are not making fun of me, are you?

I will suppose that not and will respond. You mean put it this way (\sqrt{n+1})(\sqrt{n+1}) ? No you can't use FOIL because both terms are under one root. You could have used FOIL if it would have been (\sqrt{n}+1)^2 =(\sqrt{n}+1)(\sqrt{n}+1) , but even in that case it would be more logical to use perfect square binomial identity - (a+b)^2 = a^2+2ab+b^2 ( Note : you really have to know basic polynomial identities. Make sure you know the answer to these - (a \pm b)^2 = ? a^2-b^2=? and maybe a^3 \pm b^3=? (a \pm b)^3 =? ) BUT in your case, as I said (\sqrt{n+1})^2 = ((n+1)^{1/2})^2 Note the brackets. You can't use FOIL because of the priority of the operations first you have too take the root (put to the 1/2 power) then square (multiply the brackets).... To tickle your mind (or maybe to confuse you, so may disregard this part) I could abstractly show this in terms of a function of higher order... like f(x) = n+1, g(x) = \sqrt{x}, g(f(x)) = \sqrt{f(x)}; agh never mind I will make it overly complicated.. I tried to show that while solving smth it really helps me to make an abstraction i.e. treat some polynomial as one unknown variable to see what operations I have to do with it... @ramiy make sure you don't confuse this (\sqrt{n+1})^2 with \sqrt{(n+1)^2}

Solve this root

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