In the following equation, n is an integer greater than 1 - OCR - GCSE Maths - Question 20 - 2018 - Paper 1

Given k = 64, we use the same equation:

$\left( \sqrt{2} \right)^n = 64 \sqrt{2}$

We can express 64 as a power of 2:

$64 = 2^6$

So the equation becomes:

$\left( \sqrt{2} \right)^n = 2^6 \sqrt{2} = 2^{6 + 1/2} = 2^{6.5}$

Thus, we equate exponents:

$\frac{n}{2} = 6.5$

Multiplying both sides by 2 gives:

$n = 13.$

To simplify $\frac{14}{3 - \sqrt{2}}$, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator:

$\frac{14(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}.$

The denominator simplifies as follows:

$(3 - \sqrt{2})(3 + \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7.$

Now the expression becomes:

$\frac{14(3 + \sqrt{2})}{7} = 2(3 + \sqrt{2}) = 6 + 2\sqrt{2}.$

Thus, we have shown that:

$\frac{14}{3 - \sqrt{2}} = 6 + 2\sqrt{2}.$