Question

Value of $\left(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}+\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}+\frac{\sqrt{6}}{(\sqrt{3}-\sqrt{2})+...}}}\right)^2$ is

Answer 1

2

Answer 2

Let the given expression inside the square be $y$.
The expression is
$y = \frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}+\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}+\frac{\sqrt{6}}{(\sqrt{3}-\sqrt{2})+...}}}$
This is a continued fraction. Let $a = \sqrt{3}-\sqrt{2}$. The expression can be written as
$y = \frac{\sqrt{6}}{a + \frac{\sqrt{6}}{a + \frac{\sqrt{6}}{a + ...}}}$
The structure of the continued fraction implies that the part starting from the second term in the denominator is the same as the original expression $y$.
So, we can write the equation:
$y = \frac{\sqrt{6}}{a + y}$
To solve for $y$, we multiply both sides by $(a+y)$:
$y(a+y) = \sqrt{6}$
$ay + y^2 = \sqrt{6}$
Rearranging the terms, we get a quadratic equation in $y$:
$y^2 + ay - \sqrt{6} = 0$
Substitute the value of $a = \sqrt{3}-\sqrt{2}$:
$y^2 + (\sqrt{3}-\sqrt{2})y - \sqrt{6} = 0$
We can solve this quadratic equation for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where in this case, the variable is $y$, and the coefficients are $1$, $(\sqrt{3}-\sqrt{2})$, and $-\sqrt{6}$.
Here, $a_{quad} = 1$, $b_{quad} = \sqrt{3}-\sqrt{2}$, $c_{quad} = -\sqrt{6}$.
$y = \frac{-(\sqrt{3}-\sqrt{2}) \pm \sqrt{(\sqrt{3}-\sqrt{2})^2 - 4(1)(-\sqrt{6})}}{2(1)}$
First, calculate the term under the square root:
$(\sqrt{3}-\sqrt{2})^2 - 4(-\sqrt{6}) = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 + 4\sqrt{6}$
$= 3 - 2\sqrt{6} + 2 + 4\sqrt{6}$
$= 5 + 2\sqrt{6}$
Now, we need to find the square root of $5 + 2\sqrt{6}$. We look for two numbers whose sum is 5 and product is 6. These numbers are 3 and 2.
$\sqrt{5 + 2\sqrt{6}} = \sqrt{3 + 2 + 2\sqrt{3}\sqrt{2}} = \sqrt{(\sqrt{3}+\sqrt{2})^2} = \sqrt{3}+\sqrt{2}$ (since $\sqrt{3}+\sqrt{2}$ is positive).
Substitute this back into the quadratic formula for $y$:
$y = \frac{-(\sqrt{3}-\sqrt{2}) \pm (\sqrt{3}+\sqrt{2})}{2}$
This gives two possible values for $y$:
$y_1 = \frac{-(\sqrt{3}-\sqrt{2}) + (\sqrt{3}+\sqrt{2})}{2} = \frac{-\sqrt{3}+\sqrt{2}+\sqrt{3}+\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$
$y_2 = \frac{-(\sqrt{3}-\sqrt{2}) - (\sqrt{3}+\sqrt{2})}{2} = \frac{-\sqrt{3}+\sqrt{2}-\sqrt{3}-\sqrt{2}}{2} = \frac{-2\sqrt{3}}{2} = -\sqrt{3}$

The given expression is a continued fraction of the form $\frac{p}{q + \frac{p}{q + ...}}$ where $p = \sqrt{6} > 0$ and $q = \sqrt{3}-\sqrt{2} = \frac{1}{\sqrt{3}+\sqrt{2}} > 0$. A continued fraction of this form with positive terms converges to a positive value. Therefore, the value of $y$ must be positive.
Comparing the two solutions, $y_1 = \sqrt{2}$ is positive, while $y_2 = -\sqrt{3}$ is negative.
Thus, the correct value for $y$ is $\sqrt{2}$.

The question asks for the value of the square of the given expression, which is $y^2$.
$y^2 = (\sqrt{2})^2 = 2$