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If `y=sqrtX` , what is the value of `y^25 + 121 - y^50` ?

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marrr | (Level 1) eNoter

Posted July 16, 2013 at 1:29 AM via web

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If `y=sqrtX` , what is the value of `y^25 + 121 - y^50` ?

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mvcdc | Student, Graduate | (Level 1) Associate Educator

Posted July 16, 2013 at 2:00 AM (Answer #1)

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Given that `y = sqrt(x)` , we want to calculate the value of `y^25 + 121 - y^50` , in terms of `x` .``

 

In that case, we simply substitute the value of `y` to the expression:

Note that `y = sqrt(x) = x^(1/2)` . 

Hence:

`y^25 + 121 - y^50 = (x^(1/2))^25 + 121 - (x^(1/2))^50`

Using laws of exponents we get the following expression:

`x^(25/2) + 121 - x^(50/2)`

which then simplifies to:

`x^(25/2) + 121 - x^25`

or

`sqrt(x^25) + 121 - x^25`

or

`(sqrt(x))^25 + 121 - x^25`

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