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# Algebra

Find the integer closest to $$\frac { 1 }{ \sqrt [ 4 ]{ 5^{ 4 }+1 } -\sqrt [ 4 ]{ 5^{ 4 }-1 } }$$

I wanted to post this as a question but found multiple answers Can anyone help me with this question? I think the answer might be $$1$$ OR $$0$$ OR $$250$$ which one is correct?

Note by Ankit Vijay
3 years, 6 months ago

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Consider $$\LARGE \frac {1}{ \sqrt[4]{A} - \sqrt[4]{B} }$$

Multiply numerator and denominator by $$\large \sqrt[4]{A} + \sqrt[4]{B}$$, apply difference of squares: $$x^2 - y^2 = (x-y)(x+y)$$, we have

$\LARGE \frac { \sqrt[4]{A} + \sqrt[4]{B} }{ \sqrt[2]{A} - \sqrt[2]{B} }$

Now this time, multiply numerator and denominator by $$\large \sqrt[2]{A} + \sqrt[2]{B}$$

$\LARGE \frac { \left ( \sqrt[4]{A} + \sqrt[4]{B} \right ) \left ( \sqrt[2]{A} + \sqrt[2]{B} \right ) }{ A-B }$

Now set $$A = 5^4 + 1, B = 5^4 - 1$$, the denominator equals to $$2$$

And because $$A \approx 5^4, B \approx 5^4$$, then $$\large \sqrt[4]{A} \approx 5, \sqrt[4]{B} \approx 5, \sqrt[2]{A} \approx 25, \sqrt[2]{B} \approx 25$$

So the expression is appproximately close to $$\frac {(5+5)(25+25)}{2} = \boxed{250}$$

Alternatively, you can use binomial expansion, for $$|x| <1$$, we have $$(1 + x)^n \approx 1 + nx$$

Because $$\large \sqrt[4]{5^4 + 1} = \sqrt[4]{5^4 \left ( 1 + \frac {1}{5^4} \right ) } = 5 \cdot \sqrt[4]{ 1 + \frac {1}{5^4} }$$, likewise $$\large \sqrt[4]{5^4 - 1} = 5 \cdot \sqrt[4]{ 1 - \frac {1}{5^4} }$$

So the expression equals to $$\large \left ( 5(1 + x)^n - 5(1-x)^n \right )^{-1}$$

Substitution of $$\large x = \frac {1}{5^4}, n = \frac {1}{4}$$ gives $$\large \frac {1}{10nx} = \boxed{250}$$

- 3 years, 6 months ago

- 3 years, 6 months ago

are you nuts??

- 2 years, 10 months ago