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If we write \(\sqrt{2}+\frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}\) in the form \(\dfrac{a\sqrt{2} + b\sqrt{3}}{c}\) such that \(a\)\(b\), and \(c\) are positive integers and \(c\) is as small as possible, then what is \(a + b + c\)?

Logic  Sep 23, 2018
 #1
avatar
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Simplify the following:
sqrt(2) + 1/sqrt(2) + sqrt(3) + 1/sqrt(3)

Rationalize the denominator. 1/sqrt(2) = 1/sqrt(2)×(2^(1 - 1/2))/(2^(1 - 1/2)) = (2^(1 - 1/2))/2:
sqrt(2) + (2^(1 - 1/2))/2 + sqrt(3) + 1/sqrt(3)

 

Combine powers. (2^(1 - 1/2))/2 = 2^((1 - 1/2) - 1):
sqrt(2) + 2^((1 - 1/2) - 1) + sqrt(3) + 1/sqrt(3)

Put 1 - 1/2 over the common denominator 2. 1 - 1/2 = 2/2 - 1/2:
sqrt(2) + 2^(2/2 - 1/2 - 1) + sqrt(3) + 1/sqrt(3)

2/2 - 1/2 = (2 - 1)/2:
sqrt(2) + 2^((2 - 1)/2 - 1) + sqrt(3) + 1/sqrt(3)

2 - 1 = 1:
sqrt(2) + 2^(1/2 - 1) + sqrt(3) + 1/sqrt(3)

 

Put 1/2 - 1 over the common denominator 2. 1/2 - 1 = 1/2 - 2/2:
sqrt(2) + 2^(1/2 - 2/2) + sqrt(3) + 1/sqrt(3)

1/2 - 2/2 = (1 - 2)/2:
sqrt(2) + 2^((1 - 2)/2) + sqrt(3) + 1/sqrt(3)

1 - 2 = -1:
sqrt(2) + 2^((-1)/2) + sqrt(3) + 1/sqrt(3)

 

Rationalize the denominator. 1/sqrt(2) = 1/sqrt(2)×(sqrt(2))/(sqrt(2)) = (sqrt(2))/2:
sqrt(2) + (sqrt(2))/2 + sqrt(3) + 1/sqrt(3)

Rationalize the denominator. 1/sqrt(3) = 1/sqrt(3)×(sqrt(3))/(sqrt(3)) = (sqrt(3))/3:
sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3

 

Put each term in sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3 over the common denominator 6: sqrt(2) + (sqrt(2))/2 + sqrt(3) + (sqrt(3))/3 = (6 sqrt(2))/6 + (3 sqrt(2))/6 + (6 sqrt(3))/6 + (2 sqrt(3))/6:
(6 sqrt(2))/6 + (3 sqrt(2))/6 + (6 sqrt(3))/6 + (2 sqrt(3))/6

(6 sqrt(2))/6 + (3 sqrt(2))/6 + (6 sqrt(3))/6 + (2 sqrt(3))/6 = (6 sqrt(2) + 3 sqrt(2) + 6 sqrt(3) + 2 sqrt(3))/6:
(6 sqrt(2) + 3 sqrt(2) + 6 sqrt(3) + 2 sqrt(3))/6

Add like terms. 6 sqrt(2) + 3 sqrt(2) + 6 sqrt(3) + 2 sqrt(3) = 9 sqrt(2) + 8 sqrt(3):

(9 sqrt(2) + 8 sqrt(3))/6 = a =9, b=8, c=6: a + b + c =9 + 8 + 6 = 23

Guest Sep 24, 2018
edited by Guest  Sep 24, 2018
 #2
avatar+13084 
+2

Here is my hand  scrawled answer:

 

ElectricPavlov  Sep 24, 2018

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