Question

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

Answer 1

Answer:

A+B+C+D = 13

Step-by-step explanation:

The given expression is:

$\dfrac{1+\sqrt{2}}{2+\sqrt{3}}$

We have to simply it and express it in the form of:

$A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})$

Multiply and divide the given expression with $2-\sqrt 3$:

$\dfrac{1+\sqrt{2}}{2+\sqrt{3}} \times \dfrac{2-\sqrt 3}{2-\sqrt 3}\\\Rightarrow \dfrac{(1+\sqrt{2}) \times (2-\sqrt 3)}{(2+\sqrt{3})\times (2-\sqrt 3)}\\\Rightarrow \dfrac{2+2\sqrt2-\sqrt3-\sqrt6}{2^2-(\sqrt{3})^2}\\\Rightarrow \dfrac{2+2\sqrt2-\sqrt3-\sqrt6}{4-3}\\\Rightarrow \dfrac{2(1+\sqrt2)-(\sqrt3+\sqrt6)}{1}\\\Rightarrow 2(1+\sqrt2)-(\sqrt3+\sqrt6)$

It is the simplified form of given expression.

Formula used:

$(a+b)(a-b) = a^{2} -b^{2}$

Comparing the simplified expression with $A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})$

$2(1+\sqrt2)-(\sqrt3+\sqrt6)=A(1+\sqrt{B})-(\sqrt{C}+\sqrt{D})\\\Rightarrow A =2, B=2, C=3\ and\ D=6$

So, value of

$A+B+C+D = 2+2+3+6 = 13$