Negligible radicals

Algebra Level 3

$\large {\frac { { (4+\sqrt { 15 } ) }^{ \frac { 3 }{ 2 } }+{ (4-\sqrt { 15 } ) }^{ \frac { 3 }{ 2 } } }{ { (6+\sqrt { 35 } ) }^{ \frac { 3 }{ 2 } }-{ (6-\sqrt { 35 } ) }^{ \frac { 3 }{ 2 } } } }$

If the value of the expression above equals to $\frac a b$ for coprime positive integers $a,b$ . What is the value of $a+b$ ?

This problem is a part of the set 1+2=3

13 20 15 9

3 solutions

Vermouth Xu
Apr 9, 2015

A tip: $4+\sqrt { 15 } =\quad \frac { 8+2\sqrt { 15 } }{ 2 } =\frac { { (\sqrt { 3 } +\sqrt { 5 } ) }^{ 2 } }{ 2 }$ Okay, you should know how to work this out.

Vibha Sharma
Apr 7, 2015

Anna Eldrid
Apr 11, 2015

Clue is the the title: "negligible", hence I have decided to treat them as such, and use approximations rather than calculating the whole lot. (I wonder if anyone can help me make my intuitive solution more formal). Here is my reasoning:

Since sqrt(16) is 4, sqrt(15) must be 'approaching 4', making the first term in the numerator approximately equal to 8^(3/2), whilst the second term is approximately equal to 0.

In the denominator we have the same as issue as sqrt(35) is close to sqrt(36) which is 6, hence the first therm of the denominator is close to 12^(3/2), whilst the second term gets close to 0.

So the fraction approximately equals: ((8^(3/2))/((12^(3/2)), which is the same ratio as 8/12, so the solution is 8+12= 20.

Negligible radicals

This problem is a part of the set 1+2=3

3 solutions

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