If M = square root 4 + cube root 4 + fourth root of 4, then

[Guest]

If M = square root 4 + cube root 4 + fourth root of 4, then the value of M is:

a) less than 3
b) equal to 3
c) between 3 and 4
d) equal to 4
e) greater than 4 - answer

Clearly if I had a calculator this would be easy. Am I just supposed to have memorized these numbers or is there a logical way to deal with this problem? Thanks!

[Jeff]

This is an estimation problem, you don't need to know precise values for the cube and fourth root of 4 - you just need to realize that they are each between 1 and 2. Here's why: The square root of 4 is 2. So clearly the cube and fourth root must be smaller than the square root, i.e. less than 2. However they must be greater than 1 - positive numbers less than 1 (say .75) get smaller as they are multiplied together and you need to "build-up" to 4.

So then you're left with sqrrt(4)+cubert(4)+4throot(4) = 2 + something between 1 and 2+ somthing between 1 and 2 so it must be greater than 4.

Jeff

[Guest]

Seriously, Jeff, you make this all sound so easy. :-)
Thanks. Your online explanations for this problem - as well as everyone else's problems - have been tremendously helpful.

[steph]

thanks jeff!

[RonPurewal]

well played.

there are a couple of other threads on this problem floating around, but they're hard to search (not a lot of keywords in this problem statement).

[Guest]

sqrt (4) can be minus two as well. Also fourth root of 4 can also be negative. In which case the equation cannot be greater than 4. Hence it cannot be concretely said that it will be greater than 4.

[RonPurewal]

sqrt (4) can be minus two as well.

wrong.

Also fourth root of 4 can also be negative.

also wrong.

here's the deal: you're talking about two different things.

when you have an OPERATOR SYMBOL, such as "√", that symbol will produce exactly ONE value.
for instance, √9 is just 3.
|-5| is just 5.
and so on.
notice, in these cases, that you're merely simplifying expressions; you are NOT solving EQUATIONS. in other words, there's no "=" sign in the middle of what you're dealing with; you're not finding the solutions of anything. just simplifying.

when you solve an EQUATION, there may well be more than one solution.
if you say x^2 = 9, there are two solutions: x = 3 and x = -3.
notice that i can say this without speaking of the "√" symbol at all!

indeed, the "√" symbol is DEFINED to refer only to the positive root. it is NOT defined to refer to either of the 2 solutions to the above equation.

same goes for 4th root, 6th root, and all other even roots.

hth.

[priti2000]

I represented the roots as fractions and then factored out, so:
sqrt(4) + cube root(4) + fourth root (4) =
2 + 4^1/3 + 4^1/4 =
3 + 4^1/3

which is greater than 4

[atul.prasad]

I represented the roots as fractions and then factored out, so:
sqrt(4) + cube root(4) + fourth root (4) =
2 + 4^1/3 + 4^1/4 =
3 + 4^1/3

which is greater than 4

What did you factor out?

One key thing to note is that nth root of any positive number greater than 1, HAS to be greater than 1.

this is because nth root of 1 is 1, hence the nth root of any number > 1 has to be > 1

as an example, lets prove by contradiction=>
let x = 5^(1/29)
or x^29 = 5..
now if x is < 1 , x^y WILL be < 1 for all values of y>=1
Hence for x^29 to be equal to 5, x cannot be < 1

Hence it follows that for our question,
we have sqrt(4) + cube rt(4) + 4th root(4)
2+ (a number>1) + (a number > 1)

= a number greater than 4

[jnelson0612]

Nice commentary everyone--thank you!

[aseem83]

Guys, just to add on here:

If 4(or any +ve number) raise to power 0 is = 1; then obviously any other root level(1/3,1/4,1/545) would be greater than, although by a miniscule, 1. Hence this value would always be more than 4!!

[RonPurewal]

Guys, just to add on here:

If 4(or any +ve number) raise to power 0 is = 1; then obviously any other root level(1/3,1/4,1/545) would be greater than, although by a miniscule, 1. Hence this value would always be more than 4!!

yes, this is another good way to remember that 4^fraction, or (any other number greater than 1)^fraction, is greater than 1.

it's also a good way to remember that, say, (1/2)^fraction is less than 1.

[prashant.ranjan]

The same can be said with 1 too. Such as 1 ^ (1/600) is still 1. So any root of a positive integer greater than 1 will be greater than 1.
--------
Algebraically 4 ^ (1/4) is nothing but 2 ^ (2*1/4) i.e. sqrt(2) = 1.414
So cube root of 4 will be greater than 1.414. Adding all of them will yield some integer greater than 4.

Thanks
Prashant

[RonPurewal]

The same can be said with 1 too. Such as 1 ^ (1/600) is still 1. So any root of a positive integer greater than 1 will be greater than 1.
--------
Algebraically 4 ^ (1/4) is nothing but 2 ^ (2*1/4) i.e. sqrt(2) = 1.414
So cube root of 4 will be greater than 1.414. Adding all of them will yield some integer greater than 4.

Thanks
Prashant

these are good insights.

at the same time, prashant, you should make sure that you can also, if necessary, get the much simpler fact here -- namely, the idea that all of these roots are greater than 1.
your statements here clearly demonstrate a greater-than-normal mathematical intuition -- but the flipside is that you might get caught up in unnecessary complexity. watch out for that.
basically, keep things as simple as you can, without missing key points.

[harvgill]

Here's the most elegant solution to this problem. You first raise 4 to fraction powers to represent the roots since sqrt of 4 = 4^1/2, then you add the fractional powers, then you convert 4 to 2^2, and multiply the powers to arrive at 2^2.xx


the expression could be rewritten as 4^1/2 + 4^1/3 + 4^1/4

= 4^(1/2 + 1/3 + 1/4)
= 4^(13/12)
= (2^2)^13/12
= 2^(2*13/12)
=2^(13/6)
=2^2.xxx

2 raised to any power greater than 2 will render in value of M greater than 4.