Algebra: Find $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$ using $(y-12)=4\sqrt{5}$

The problem asks for the value of the expression $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$, given the equation $(y-12) = 4\sqrt{5}$.

Step 1: Determine the value of $y-3$

Start with the given equation:

$y - 12 = 4\sqrt{5}$

To find $y$, add 12 to both sides:

$y = 12 + 4\sqrt{5}$

Now, calculate $y-3$:

$y - 3 = (12 + 4\sqrt{5}) - 3$

$y - 3 = 9 + 4\sqrt{5}$

Step 2: Simplify the term $\sqrt{y-3}$

We need to calculate $\sqrt{9 + 4\sqrt{5}}$. This involves simplifying a nested radical.

Rewrite the expression in the form $\sqrt{a+b+2\sqrt{ab}}$:

$4\sqrt{5} = 2 \times 2\sqrt{5} = 2\sqrt{4 \times 5} = 2\sqrt{20}$

So, the expression becomes:

$\sqrt{9 + 2\sqrt{20}}$

Look for two numbers that sum to 9 and multiply to 20. These numbers are 5 and 4.

$\sqrt{9 + 2\sqrt{20}} = \sqrt{(5 + 4) + 2\sqrt{5 \times 4}} = \sqrt{(\sqrt{5} + \sqrt{4})^2}$

Therefore, $\sqrt{y-3}$ simplifies to:

$\sqrt{y-3} = \sqrt{5} + \sqrt{4} = \sqrt{5} + 2$

Step 3: Evaluate the target expression

Substitute $\sqrt{y-3} = \sqrt{5} + 2$ into the expression $\sqrt{y-3} - \frac{1}{\sqrt{y-3}}$:

$(\sqrt{5} + 2) - \frac{1}{(\sqrt{5} + 2)}$

Rationalize the denominator of the fraction $\frac{1}{(\sqrt{5} + 2)}$:

$\frac{1}{\sqrt{5} + 2} = \frac{1}{(\sqrt{5} + 2)} \times \frac{(\sqrt{5} - 2)}{(\sqrt{5} - 2)} = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - 2^2}$

$= \frac{\sqrt{5} - 2}{5 - 4} = \frac{\sqrt{5} - 2}{1} = \sqrt{5} - 2$

Now substitute this back into the expression:

$(\sqrt{5} + 2) - (\sqrt{5} - 2)$

$= \sqrt{5} + 2 - \sqrt{5} + 2$

$= 4$

The value of the expression is 4.