{1/(4-√t15)}^2=

Multiplying, the numerator and denominator by \((4+\sqrt{15})\), applying \(a^2-b^2\) and \((a+b)^2\) formulae.

\(=\frac{{16+8\sqrt{15}+15}}{16-15}\)

\(31+8\sqrt{15}\)

The answer is \(C\)

\((\frac{1}{{4-√15}})^2 \)

\(=(\frac{(4+√15)}{(4-√15)(4+√15)})^2 \)

\(=(\frac{16+8√15+15}{16-15}) \)

\(= 31+8√15\), Answer is (C)

Strategy: For this question, we can either expand \((\frac{1}{4-\sqrt{15}})^2\) and then "fix" the resulting denominator, or " fix" the denominator first and then expand.
Let's fix the denominator first

Take: \(\frac{1}{4-\sqrt{15}}\)

Multiply numerator and denominator by \(4+\sqrt{15}\), which is the complement of \(4-\sqrt{15}\)

When we do this we get: \(\frac{1}{4-\sqrt{15}} = \frac{(1)(4+\sqrt{15})}{(4-\sqrt{15})(4+\sqrt{15})}= \frac{4+\sqrt{15}}{16 - 15}= \frac{4+\sqrt{15}}{1} = 4+\sqrt{15}\)

Substitute this "fixed" expression into the original expression to get: \((\frac{1}{4-\sqrt{15}})^2 = (4+\sqrt{15})^2 = (4+\sqrt{15})(4+\sqrt{15}) = 16 + 4\sqrt{15} + 4\sqrt{15} + (\sqrt{15})^2 = 31 + 8\sqrt{15}\)

Answer: C

\(\left[ \begin{array}{cc|r} \dfrac{1}{4-\sqrt{15}} \end{array} \right]^2\)

This problem is based on concept of Roots called as Conjugate Roots.

In these kind of problem to simplify the expression (and the denominator) we multiply the numerator and the denominator with the Conjugate of the denominator

Conjugate of \(4-\sqrt{15}\) is \(4+\sqrt{15}\) (Same expression with sign of the square root term reversed)

=> \((\frac{1}{4-\sqrt{15}})^2\) = \((\frac{1}{4-\sqrt{15}} * \frac{4+\sqrt{15}}{4+\sqrt{15}})^2\)
= \((\frac{4+\sqrt{15}}{(4-\sqrt{15})*(4+\sqrt{15})})^2\)

Using (a-b)*(a+b) = \(a^2 - b^2\) in the denominator where a = 4 and b = \(\sqrt{15}\), we get

= \((\frac{4+\sqrt{15}}{(4^2)-15})^2\)
= \((\frac{4+\sqrt{15}}{16-15})^2\)
= \((\frac{4+\sqrt{15}}{1})^2\)
= \((4+\sqrt{15})^2\)
= \(4^2 + 2*4*\sqrt{15} + 15\)
= 31 + 8\(\sqrt{15}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Rationalize Roots