how to evaluate the square root of n + the square root of the square root of n plus the square root of the square root of the square root of n.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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You solve an equation or inequality, so you are not being asked to set the expression equal to zero since you are asked to evaluate.

To evaluate an expression, you should be asked to evaluate at some input.

For instance, to evaluate `sqrt(n)+sqrt(sqrt(n))+sqrt(sqrt(sqrt(n)))` at 256 we have:

`sqrt(256)=16`

`sqrt(sqrt(n))=sqrt(sqrt(256))=sqrt(16)=4`

`sqrt(sqrt(sqrt(256)))=sqrt(sqrt(16))=sqrt(4)=2`

So the sum is 22.

You would follow the same type of procedure for whatever input value you are given.

** With rational exponents you can write the sum as `n^(1/2)+n^(1/4)+n^(1/8)` and then evaluate at the given n**

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The request of the problem is vague, hence, if you need to evaluate for n the equation `sqrt n + sqrt(sqrt n) + sqrt(sqrt(sqrt n)) = 0` , you should use the following substitution such that:

`sqrt n = t`

`t + sqrt t + sqrt(sqrt t) = 0`

You need to move `sqrt(sqrt t)` to the right side such that:

`t + sqrt t = -sqrt(sqrt t)`

You need to raise to square both sides, such that:

`(t + sqrt t)^2 = (-sqrt(sqrt t))^2`

`t^2 + 2tsqrt t + t = sqrt t`

`t^2 + 2tsqrt t + t - sqrt t = 0`

`t^2 + t = sqrt t - 2t*sqrt t`

Raising to square both sides yields:

`t^4 + 2t^3 + t^2 = t - 4t^2 + 4t^3`

`t^4 + 2t^3 + t^2 - t + 4t^2 - 4t^3 = 0`

`t^4 - 2t^3 + 5t^2 - t = 0`

You need to factor out t such that:

`t(t^3 - 2t^2 + 5t - 1) = 0`

Notice that the equation `t^4 - 2t^3 + 5t^2 - t = 0` has the real roots `t = 0` and `t = 0.2` and two complex conjugate solutions.

You need to substitute back `sqrt n` for ` t` such that:

`sqrt n = 0 => n = 0`

`sqrt n = 0.2 => n = 0.2^2 => n = 0.04`

Hence, evaluating the real solutions to the equation `sqrt n + sqrt(sqrt n) + sqrt(sqrt(sqrt n)) = 0` yields `n = 0` and `n = 0.04.`

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