\text{If }(y-12)=4\sqrt{5}\text{, then find the value of }\sqrt{y-3}-\frac{1}{\sqrt{y-3}}\text{.}

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The correct answer is 4.

Step 1: Simplify the expression inside the square root.

We are given the equation:

y-12=4\sqrt{5}

We need to find the value of \sqrt{y-3}-\frac{1}{\sqrt{y-3}}. Let us first express y-3 in terms of the given value:

y-3=(y-12)+9

Substitute y-12=4\sqrt{5} into the equation:

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

Step 2: Express y-3 as a perfect square.

Notice that we can rewrite 9+4\sqrt{5} in the form of {(a+b)}^2=a^2+b^2+2ab:

9+4\sqrt{5}=5+4+2\cdot 2\cdot \sqrt{5}

y-3={(\sqrt{5})}^2+2^2+2(\sqrt{5})(2)={(\sqrt{5}+2)}^2

Taking square root on both sides:

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

Step 3: Calculate the reciprocal.

Now, rationalizing the denominator for \frac{1}{\sqrt{y-3}}:

\frac{1}{\sqrt{y-3}}=\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)}

\frac{1}{\sqrt{y-3}}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2

Step 4: Compute the final value.

Subtracting the reciprocal from \sqrt{y-3}:

\sqrt{y-3}-\frac{1}{\sqrt{y-3}}=(\sqrt{5}+2)-(\sqrt{5}-2)

=\sqrt{5}+2-\sqrt{5}+2=4