Indices, roots, standard form and exact calculation Revision Notes

What are Exact Calculations?

Exact calculations mean working with numbers without rounding them. This keeps the answer perfectly precise. Instead of converting a fraction like 1/3 to a decimal like 0.33, you keep it as 1/3. For irrational numbers like pi (π) or the square root of 2 (√2), you leave them in their symbolic form rather than using rounded decimal approximations. This is important because any rounding, even minor, can lead to inaccuracies in the final answer, especially in problems with many steps.

Why are Exact Calculations Important?

Exact calculations are vital for precision. Rounding numbers too early in a calculation can lead to a 'rounding error' where the final answer is not as accurate as it should be. This can be critical in fields like engineering, science, or finance where small errors can have large consequences. By using exact values, you ensure that every step of your calculation is perfectly accurate, leading to a perfectly accurate final answer if one exists.

Methods for Exact Calculations

To perform exact calculations, you need to be comfortable with several mathematical techniques. These include:

1. Fractions: Keep numbers as fractions throughout calculations (e.g., adding, subtracting, multiplying, dividing fractions). Convert mixed numbers to improper fractions when performing operations.
2. Surds: Simplify and manipulate square roots (like √8 = 2√2) without converting them to decimals. Operations with surds (addition, subtraction, multiplication, division) should maintain their exact form.
3. Constants: Use symbolic forms for constants like π (leave as π) rather than 3.14 or 22/7, unless the question specifically requires an approximation.
4. Algebraic Expressions: When dealing with algebraic terms, maintain all coefficients and constants in their exact form.

When to Use Exact Calculations

Always aim for exact calculations unless the question explicitly asks for a rounded answer (e.g., 'give your answer to two decimal places' or 'give your answer to three significant figures'). If a question asks for the 'exact value' or does not specify a degree of accuracy, your answer should be in its precise form, often as a fraction, surd, or in terms of π. Practicing these methods ensures you can always provide the most accurate answer possible.

Working with Surds

A surd is a square root that cannot be simplified to a whole number, e.g., √2, √3, √5. To perform exact calculations with surds, remember key rules: √(ab) = √ a × √ b and √(a/b) = √ a / √ b. When adding or subtracting surds, they must have the same root part to be combined, similar to combining like terms in algebra (e.g., 2√3 + 5√3 = 7√3). Always simplify surds by factoring out perfect squares, for instance, √12 = √(4 × 3) = √4 × √3 = 2√3. This simplification is often the first step in exact calculations involving surds.

Rationalising Denominators

Rationalising a denominator means changing an expression so that there is no surd in the denominator. If the denominator is a single surd, like 1/√ a, multiply both the numerator and denominator by √ a. For example, 5/√3 = (5 × √3) / (√3 × √3) = 5√3 / 3. If the denominator is of the form a + √ b or a - √ b, multiply the numerator and denominator by its conjugate. The conjugate of a + √ b is a - √ b, and vice versa. This uses the difference of two squares property (A+B)(A-B) = A² - B², which eliminates the surd. For example, to rationalise 1/(2 + √3), multiply by (2 - √3)/(2 - √3) to get (2 - √3) / (4 - 3) = 2 - √3.

Exact Values with Pi (π)

Many geometrical calculations, such as the area of a circle or the volume of a sphere, involve the constant π. When asked for an exact answer, leave π in your final expression rather than substituting an approximate decimal value like 3.14 or 22/7. For example, if a circle has a radius of 5 cm, its exact area would be π r² = π (5²) = 25π cm². Similarly, volume calculations for cylinders or cones should retain π for exact answers. Always ensure that the units are correctly stated when giving an exact answer involving π.

Solving Equations Exactly

Higher-level problems often require solving equations to find exact solutions. For quadratic equations that don't factorise easily, use the quadratic formula, x = [-b ± √(b² - 4ac)] / 2a. The solutions may involve surds, and these should be simplified and left in their exact form. For example, if x² - 6x + 7 = 0, using the formula gives x = [6 ± √(36 - 28)] / 2 = [6 ± √8] / 2 = [6 ± 2√2] / 2 = 3 ± √2. Avoid decimal approximations unless specifically requested, as this will lead to a loss of marks in exact calculation questions.

Simplifying Surds

To simplify a surd of the form \(\sqrt{a}\), identify the largest perfect square factor of 'a'. Let this factor be \(k^2\), so \(a = k^2 \times m\). Then, \(\sqrt{a} = \sqrt{k^2 \times m} = \sqrt{k^2} \times \sqrt{m} = k\sqrt{m}\). For example, to simplify \(\sqrt{72}\):

1. Find the largest perfect square factor of 72, which is 36.
2. Write \(72 = 36 \times 2\).
3. So, \(\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}\).

This process aims to reduce the number under the root sign to its simplest form.

Operations with Surds

Surds can be added and subtracted only if they are 'like surds' (have the same number under the radical sign). If they are not like surds, simplify them first. For example, \(2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}\). If you have \(\sqrt{20} + \sqrt{45}\), first simplify: \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\) and \(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\). Then, \(2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}\). For multiplication, use the rule \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\). For example, \(\sqrt{3} \times \sqrt{7} = \sqrt{21}\). For division, use \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\) or \(\sqrt{a} \div \sqrt{b} = \sqrt{\frac{a}{b}}\). For example, \(\frac{\sqrt{30}}{\sqrt{5}} = \sqrt{\frac{30}{5}} = \sqrt{6}\). Expanding brackets with surds follows the same rules as algebraic expansion, e.g., \((a+\sqrt{b})(c+\sqrt{d}) = ac + a\sqrt{d} + c\sqrt{b} + \sqrt{bd}\).

Rationalising the Denominator

To rationalise the denominator means to remove any surds from the denominator of a fraction. This is done by multiplying the numerator and denominator by an appropriate expression.

1. If the denominator is a single surd, say \(\sqrt{a}\), multiply both numerator and denominator by \(\sqrt{a}\). For example, \(\frac{x}{\sqrt{a}} = \frac{x}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} = \frac{x\sqrt{a}}{a}\).
2. If the denominator is a binomial involving a surd, such as \(a + \sqrt{b}\) or \(a - \sqrt{b}\), multiply both numerator and denominator by its conjugate. The conjugate of \(a + \sqrt{b}\) is \(a - \sqrt{b}\), and vice versa. This uses the difference of two squares identity: \((x+y)(x-y) = x^2 - y^2\). For example, \(\frac{1}{3 - \sqrt{2}} = \frac{1}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{3 + \sqrt{2}}{3^2 - (\sqrt{2})^2} = \frac{3 + \sqrt{2}}{9 - 2} = \frac{3 + \sqrt{2}}{7}\).