Surds and Indices: Simplify, Rationalise, and Calculate with Confidence

What the examiner is testing

The examiner is assessing your ability to manipulate expressions involving roots and powers, specifically simplifying surds, rationalising denominators, and applying the rules of indices. Marks are awarded for demonstrating correct algebraic steps in transforming expressions into their simplest forms or evaluating them accurately.

The method

1. Identify whether the expression contains surds, indices, or both.
2. If simplifying a surd, look for the largest square number factor of the number under the root. For example, for \( \sqrt{72} \), \( 36 \) is the largest square factor.
3. Rewrite the surd as a product of two surds, one containing the square factor. For \( \sqrt{72} \), this becomes \( \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} \).
4. Simplify the square root of the square factor. For \( \sqrt{36} \times \sqrt{2} \), this becomes \( 6\sqrt{2} \).
5. If rationalising a denominator with a single surd, multiply both the numerator and denominator by that surd. For example, for \( \frac{3}{\sqrt{5}} \), multiply by \( \frac{\sqrt{5}}{\sqrt{5}} \).
6. If rationalising a denominator with a binomial surd (e.g., \( a + \sqrt{b} \) or \( a - \sqrt{b} \)), multiply both the numerator and denominator by its conjugate. The conjugate of \( a + \sqrt{b} \) is \( a - \sqrt{b} \).
7. When applying rules of indices, ensure the bases are the same if adding or subtracting powers. If not, try to rewrite them with a common base.
8. Apply the relevant index rule: \( a^m \times a^n = a^{m+n} \), \( a^m \div a^n = a^{m-n} \), \( (a^m)^n = a^{mn} \), \( a^0 = 1 \), \( a^{-n} = \frac{1}{a^n} \), \( a^{\frac{1}{n}} = \sqrt[n]{a} \), \( a^{\frac{m}{n}} = (\sqrt[n]{a})^m \).
9. Perform any necessary multiplication or division of coefficients separately from the surd or index parts.
10. Combine like terms, ensuring surds are identical before adding or subtracting their coefficients.

Worked example

Simplify \( \frac{10}{\sqrt{2}} + \sqrt{50} \).

$$ \begin{aligned} \frac{10}{\sqrt{2}} + \sqrt{50} &= \frac{10}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{25 \times 2} \\ &= \frac{10\sqrt{2}}{2} + \sqrt{25} \times \sqrt{2} \\ &= 5\sqrt{2} + 5\sqrt{2} \\ &= 10\sqrt{2} \end{aligned} $$
Sanity check: \( \sqrt{2} \) is approximately \( 1.414 \). So \( \frac{10}{\sqrt{2}} \approx \frac{10}{1.414} \approx 7.07 \). \( \sqrt{50} \approx \sqrt{49} = 7 \). The sum is approximately \( 7.07 + 7 = 14.07 \). Our answer \( 10\sqrt{2} \approx 10 \times 1.414 = 14.14 \). The values are close, suggesting the simplification is correct.

Worked example: a harder one

Show that \( (3 + \sqrt{5})^2 - (3 - \sqrt{5})^2 \) can be written in the form \( a\sqrt{b} \).

A common first move is to expand each bracket separately:
\( (3 + \sqrt{5})^2 = 3^2 + 2(3)(\sqrt{5}) + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \)
\( (3 - \sqrt{5})^2 = 3^2 - 2(3)(\sqrt{5}) + (\sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5} \)
Then subtract:
\( (14 + 6\sqrt{5}) - (14 - 6\sqrt{5}) = 14 + 6\sqrt{5} - 14 + 6\sqrt{5} = 12\sqrt{5} \)
This works, but it's more prone to sign errors. The obvious first move is not always the most efficient or safest.

A more elegant approach recognises the difference of two squares: \( x^2 - y^2 = (x - y)(x + y) \).
Let \( x = (3 + \sqrt{5}) \) and \( y = (3 - \sqrt{5}) \).
$$ \begin{aligned} (3 + \sqrt{5})^2 - (3 - \sqrt{5})^2 &= ((3 + \sqrt{5}) - (3 - \sqrt{5}))((3 + \sqrt{5}) + (3 - \sqrt{5})) \\ &= (3 + \sqrt{5} - 3 + \sqrt{5})(3 + \sqrt{5} + 3 - \sqrt{5}) \\ &= (2\sqrt{5})(6) \\ &= 12\sqrt{5} \end{aligned} $$
This is in the form \( a\sqrt{b} \) where \( a=12 \) and \( b=5 \). This method requires fewer steps and reduces the chance of arithmetic errors.

Practice

1. Simplify \( \sqrt{75} - \sqrt{12} \).
2. Rationalise the denominator of \( \frac{6}{3 + \sqrt{3}} \).
3. Evaluate \( (27^{\frac{2}{3}})^{\frac{1}{2}} \).
4. Given that \( x = 2^{p} \) and \( y = 2^{q} \), express \( \frac{x^2 \times y}{8} \) in the form \( 2^k \).

Answers:
1. \( 3\sqrt{3} \)
2. \( 3 - \sqrt{3} \)
3. \( \sqrt{3} \)
4. Working for question 4:
$$ \begin{aligned} \frac{x^2 \times y}{8} &= \frac{(2^p)^2 \times 2^q}{2^3} \\ &= \frac{2^{2p} \times 2^q}{2^3} \\ &= \frac{2^{2p+q}}{2^3} \\ &= 2^{2p+q-3} \end{aligned} $$
So \( k = 2p+q-3 \).

The three mistakes that lose marks

1. Incorrectly simplifying surds by "splitting" non-square factors:
   Wrong: \( \sqrt{12} = \sqrt{6 \times 2} = \sqrt{6} \times \sqrt{2} \). This isn't simplified and doesn't use a square factor.
   Correct: \( \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \).
2. Applying index rules incorrectly when bases are different or operations are addition/subtraction:
   Wrong: \( 2^3 + 2^4 = 2^7 \). This is incorrect; \( 8 + 16 = 24 \), not \( 128 \). The rule \( a^m \times a^n = a^{m+n} \) applies to multiplication, not addition.
   Correct: \( 2^3 \times 2^4 = 2^{3+4} = 2^7 \).
3. Errors when rationalising binomial denominators, especially with signs:
   Wrong: Rationalising \( \frac{1}{2 + \sqrt{3}} \) by multiplying by \( \frac{2 + \sqrt{3}}{2 + \sqrt{3}} \). This results in \( \frac{2 + \sqrt{3}}{(2 + \sqrt{3})^2} \), which still has a surd in the denominator.
   Correct: Multiply by the conjugate \( \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \). This gives \( \frac{1 \times (2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \).

30-second recap

Surds are simplified by extracting square factors. Rationalising removes surds from the denominator using multiplication by the surd itself or its conjugate. Index rules apply to powers with the same base for multiplication and division, and for powers of powers.