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# Algebra

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Simplify $\frac{1+\sqrt{2}}{1+\sqrt{3}} \cdot 2$. Your solution can be converted to the form $-A-\sqrt{B}+\sqrt{C}+\sqrt{D}$, where A, B, C, and D are positive integers. What is $A+B+C+D$?

Jun 9, 2022

#1
+125695
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$$\frac{1+\sqrt{2}}{1+\sqrt{3}} \cdot 2$$

$$-A-\sqrt{B}+\sqrt{C}+\sqrt{D}$$

Multiply  num  /  den  by       1 - sqrt 3

(1 + sqrt 2) ( 1  -sqrt 3)          1 -sqrt 2 + sqrt 3 - sqrt 6          1  - sqrt 2 + sqrt 3 - sqrt 6

__________________ =      ____________________ =     _______________________

( 1 + sqrt 3) (1 - sqrt 3)                  1 -  3                                                 -2

Multiplying this result by 2  we get

- 1 ( 1 - sqrt 2  + sqrt 3  - sqrt 6 )  =

-1 - sqrt 3 + sqrt 2  +  sqrt  6

A + B + C + D   =      1  + 3  + 2  + 6   =     12

Jun 9, 2022
#2
+2665
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We have: $${1 + \sqrt2 \over 1 + \sqrt 3 } \times 2$$

First, we need to rationalize the denominator of the fraction by multiplying it by $${1 - \sqrt 3 \over 1 - \sqrt 3 }$$, which gives us $${\sqrt 2 - \sqrt 6 +1 - \sqrt 3 \over -2}$$

MUltiplying this fraction by 2 gives us: $$-1(\sqrt 2 - \sqrt 6 + 1 - \sqrt 3) = -1 - \sqrt 2 + \sqrt 3 + \sqrt 6$$

Thus, the sum is $$1 + 2 + 3 + 6 = \color{brown}\boxed{12}$$

Jun 9, 2022