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

To find n when k = 64, we start from the initial equation:

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

This can be rewritten as:

$\sqrt{2}^n = 64 \sqrt{2}$

Rearranging gives:

$\left(\sqrt{2}\right)^{n-1} = 64$

Expressing 64 as a power of 2, we have:

$64 = 2^6$

Since $\sqrt{2} = 2^{1/2}$, we can write the left side as:

$\left(2^{1/2}\right)^{n-1} = 2^{(n-1)/2}$

Therefore, we equate the exponents:

$\frac{n-1}{2} = 6$

Solving for n, we multiply both sides by 2:

$n - 1 = 12$

Thus,

$n = 13.$

To express ( \frac{14}{3 - \sqrt{2}} ) in the form a + b \sqrt{2}, we will rationalize the denominator. We multiply the numerator and the denominator by the conjugate of the denominator:

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

Calculating the denominator using the difference of squares:

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

Thus, we have:

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

This simplifies to:

$\frac{14 \cdot 3}{7} + \frac{14 \cdot \sqrt{2}}{7} = 6 + 2\sqrt{2}$

So we can express it as:

$a = 6, b = 2, \text{ leading to } 6 + 2\sqrt{2}.$