Look for the largest square number that divides 54.

\(54 = 9 \times 6\), and 9 is a square number.

A root of a product is the product of the roots:

\(\sqrt{ 54 } = \sqrt{ 9 } \times \sqrt{ 6 } = 3\sqrt{ 6 }\)

6 has no square factor left in it, so 6 is as small as it can be.

Simplify the square root first and leave the 4 where it is.

\(32 = 16 \times 2\), so \(\sqrt{ 32 } = 4\sqrt{ 2 }\)

Now multiply by the 4 in front:

\(4 \times 4\sqrt{ 2 } = 16\sqrt{ 2 }\)

Take the 7 back inside the root. A number outside a square root goes in as its square.

\(7\sqrt{ 5 } = \sqrt{ 49 \times 5 } = \sqrt{ 245 }\)

So \(n = 245\)

Check the other way: \(5\) has no square factor, so \(\sqrt{ 245 }\) really does simplify to \(7\sqrt{ 5 }\).

Substitute \(a = 3\) and \(b = 12\), then simplify. The surd in \(b\) simplifies first, because 12 is 4 times 3:

\(\sqrt{ 12 } = \sqrt{ 4 \times 3 } = 2\sqrt{ 3 }\)

The values of the possible expressions are:

\(\sqrt{ a b } = \sqrt{ 3 \times 12 } = \sqrt{ 36 } = 6\), and \(\sqrt{ a } \times \sqrt{ b } = 6\) too

\(\sqrt{ b } = 2\sqrt{ 3 }\), by the line above

\(\frac{ 12 }{ \sqrt{ 3 } } = \frac{ 12\sqrt{ 3 } }{ 3 } = 4\sqrt{ 3 }\)

\(3\sqrt{ 12 } = 3 \times 2\sqrt{ 3 } = 6\sqrt{ 3 }\)

\(\sqrt{ 3 } + \sqrt{ 12 } = \sqrt{ 3 } + 2\sqrt{ 3 } = 3\sqrt{ 3 }\)

\(12\sqrt{ 3 }\) is already as simple as it can be.

So \(\sqrt{b}\) matches \(2\sqrt{3}\), \(a\sqrt{b}\) matches \(6\sqrt{3}\) and \(\sqrt{ab}\) matches \(6\).

The hypotenuse is the longest side, so x is a shorter side: subtract the square of the known shorter side from the square of the hypotenuse.

Squaring a surd gives a whole number, so every square is an integer.

\(x^2 = 82 - 7 = 75\)

Now take the square root and pull out the largest square factor.

\(x = \sqrt{ 75 } = \sqrt{ 25 \times 3 } = 5\sqrt{ 3 }\) cm

Neither surd is in its simplest form, so simplify each one before trying to add them.

\(\sqrt{ 275 } = \sqrt{ 25 \times 11 } = 5\sqrt{ 11 }\)

\(\sqrt{ 44 } = \sqrt{ 4 \times 11 } = 2\sqrt{ 11 }\)

Both are now multiples of \(\sqrt{ 11 }\), so they collect exactly like any other like terms:

\(5\sqrt{ 11 } + 2\sqrt{ 11 } = 7\sqrt{ 11 }\)

Simplify each surd first.

\(\sqrt{ 75 } = \sqrt{ 25 \times 3 } = 5\sqrt{ 3 }\)

\(\sqrt{ 12 } = \sqrt{ 4 \times 3 } = 2\sqrt{ 3 }\)

Both are multiples of \(\sqrt{ 3 }\), so subtract the numbers in front:

\(5\sqrt{ 3 } - 2\sqrt{ 3 } = 3\sqrt{ 3 }\)

Multiply each term inside the bracket by \(\sqrt{ 3 }\).

\(\sqrt{ 3 } \times \sqrt{ 3 } = 3\), because multiplying a square root by itself gives back the number.

\(\sqrt{ 3 } \times 5 = 5\sqrt{ 3 }\)

\(\sqrt{ 3 }\left(\sqrt{ 3 } + 5\right) = 3 + 5\sqrt{ 3 }\)

Multiply out all four products.

\(2 \times 3 = 6\)

\(2 \times \left(-\sqrt{ 10 }\right) = -2\sqrt{ 10 }\)

\(\sqrt{ 10 } \times 3 = 3\sqrt{ 10 }\)

\(\sqrt{ 10 } \times \left(-\sqrt{ 10 }\right) = -10\)

Collect the whole numbers and the surds separately:

\(6 - 10 = -4\) and \(-2\sqrt{ 10 } + 3\sqrt{ 10 } = \sqrt{ 10 }\)

\(-4 +\sqrt{ 10 }\)

This is a difference of two squares, so the two surd terms cancel.

\(2 \times 2 = 4\) and \(\sqrt{ 5 } \times \sqrt{ 5 } = 5\)

\(-2\sqrt{ 5 } + 2\sqrt{ 5 } = 0\)

\(4 - 5 = -1\)

The answer is a whole number: every pair of brackets like this one rationalises, which is what makes the next question work.

Multiply the top and the bottom by \(\sqrt{ 7 }\). Multiplying by the same thing top and bottom leaves the value alone, and it clears the root from the bottom.

\(\frac{ 28 }{\sqrt{ 7 } } = \frac{ 28\sqrt{ 7 } }{\sqrt{ 7 } \times \sqrt{ 7 } } = \frac{ 28\sqrt{ 7 } }{ 7 }\)

\(28 \div 7 = 4\), so this simplifies to \(4\sqrt{ 7 }\)

Multiply the top and the bottom by \(\sqrt{ 7 }\):

\(\frac{ 2 }{\sqrt{ 7 } } = \frac{ 2\sqrt{ 7 } }{\sqrt{ 7 } \times \sqrt{ 7 } } = \frac{ 2\sqrt{ 7 } }{ 7 }\)

2 and 7 share no factor other than 1, so this will not cancel any further.

Rationalise the first term by multiplying top and bottom by \(\sqrt{ 5 }\):

\(\frac{ 7\sqrt{ 7 } }{\sqrt{ 5 } } = \frac{ 7\sqrt{ 7 }\times\sqrt{ 5 } }{ 5 } = \frac{ 7\sqrt{ 35 } }{ 5 }\)

In the second term, \(\sqrt{ 20 } = \sqrt{ 2^2 \times 5 } = 2\sqrt{ 5 }\). Rationalise by multiplying top and bottom by \(\sqrt{ 7 }\):

\(\frac{\sqrt{ 20 } }{\sqrt{ 7 } } = \frac{ 2\sqrt{ 5 }\times\sqrt{ 7 } }{ 7 } = \frac{ 2\sqrt{ 35 } }{ 7 }\)

Both terms are multiples of \(\sqrt{ 35 }\), so they can be subtracted like surds once they share a denominator. The denominators are 5 and 7, so use 35:

\(\frac{ 7\sqrt{ 35 } }{ 5 } - \frac{ 2\sqrt{ 35 } }{ 7 } = \frac{ 49\sqrt{ 35 } }{ 35 } - \frac{ 10\sqrt{ 35 } }{ 35 }\)

\(= \frac{ (49 - 10)\sqrt{ 35 } }{ 35 } = \frac{ 39\sqrt{ 35 } }{ 35 }\)

39 and 35 have no common factor, so this is in lowest terms and has the required form, with \(m = 39\), \(n = 35\) and \(k = 35\).

The top has two different surds, so simplify the larger one. \(27\) is \(9 \times 3\), and \(9\) is a square number.

\(\sqrt{ 27 } = \sqrt{ 9 } \times \sqrt{ 3 } = 3\sqrt{ 3 }\)

The top is now two like surds, which can be combined:

\(3\sqrt{ 3 } - \sqrt{ 3 } = 2\sqrt{ 3 }\)

So the fraction is \(\frac{ 2\sqrt{ 3 } }{ 2\sqrt{ 3 } }\). The \(\sqrt{ 3 }\) cancels and \(2 \div 2 = 1\).

The fraction is equal to \(1\), which is an integer.

Multiply the top and the bottom by the conjugate \(3 - \sqrt{ 7 }\). The bottom then becomes a difference of two squares, and the root disappears:

\(\left(3 + \sqrt{ 7 }\right)\left(3 - \sqrt{ 7 }\right) = 9 - 7 = 2\)

The top becomes \(2\left(3 - \sqrt{ 7 }\right)\), so

\(\frac{ 2 }{ 3 + \sqrt{ 7 } } = \frac{ 2\left(3 - \sqrt{ 7 }\right)}{ 2 }\)

\(2 \div 2 = 1\), so the answer is \(3 - \sqrt{ 7 }\)

The conjugate of \(6 - \sqrt{ 13 }\) is \(6 + \sqrt{ 13 }\). Multiply the top and the bottom by it.

\(\left(6 - \sqrt{ 13 }\right)\left(6 + \sqrt{ 13 }\right) = 36 - 13 = 23\)

\(\frac{ 46 }{ 6 - \sqrt{ 13 } } = \frac{ 46\left(6 + \sqrt{ 13 }\right)}{ 23 }\)

\(46 \div 23 = 2\), so the answer is \(12 + 2\sqrt{ 13 }\)