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How do you manipulate powers and irrational roots exactly, without a calculator?

Laws of indices for all rational exponents, surd manipulation and rationalising denominators, and the meaning of negative and fractional indices.

A focused answer to the OCR A-Level Mathematics A indices and surds content, covering the laws of indices for all rational exponents, negative and fractional powers, simplifying surds, and rationalising denominators including those of the form a plus root b.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
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What this dot point is asking

OCR wants you to apply the laws of indices for all rational exponents, interpret negative and fractional powers, simplify and combine surds, and rationalise denominators, including denominators of the form a+bca + b\sqrt{c} using the conjugate. These are non-calculator skills that underpin algebra throughout the course.

The answer

The laws of indices

For any non-zero base and rational powers:

A fractional power am/na^{m/n} means "take the nnth root, then raise to the mmth power" (the order does not matter, but rooting first usually keeps numbers small). A negative power means "take the reciprocal".

Surds

A surd is an irrational root such as 2\sqrt{2} or 53\sqrt[3]{5}. The key manipulation rules are ab=a b\sqrt{ab} = \sqrt{a}\,\sqrt{b} and ab=ab\sqrt{\tfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}. To simplify a surd, take out the largest square factor: 72=36×2=62\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}.

You can add or subtract only like surds: 35+25=553\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}, but 2+3\sqrt{2} + \sqrt{3} cannot be combined.

Rationalising the denominator

A fraction is "rationalised" when no surd appears in the denominator. For a single surd, multiply top and bottom by that surd. For a denominator a+bca + b\sqrt{c}, multiply by the conjugate a−bca - b\sqrt{c}, because (a+bc)(a−bc)=a2−b2c(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2 c is rational.

Examples in context

Combining the skills

Index and surd work appears inside almost every algebra question, for instance simplifying a derivative written with fractional powers, or tidying an answer so a marker can read it.

Solving an equation with a hidden quadratic in a power

Index laws let you spot a "hidden quadratic" when an unknown appears in an exponent, for example 4x−5(2x)+4=04^x - 5(2^x) + 4 = 0. Writing y=2xy = 2^x turns 4x=(2x)2=y24^x = (2^x)^2 = y^2, so the equation becomes the quadratic y2−5y+4=0y^2 - 5y + 4 = 0. This substitution trick recurs in the exponentials topic too.

Why exact form matters

OCR frequently demands answers "in exact form" or "in the form a+bca + b\sqrt{c}". A decimal from a calculator would lose marks where an exact surd or fraction is required, so rationalising and simplifying surds is not optional tidying but a graded skill. Leaving a surd in a denominator, or a decimal where a fraction is asked for, is treated as an incomplete answer.

Try this

Q1. Simplify 50+18\sqrt{50} + \sqrt{18}. [2 marks]

  • Cue. 52+32=825\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}.

Q2. Rationalise and simplify 63\dfrac{6}{\sqrt{3}}. [2 marks]

  • Cue. Multiply by 3/3\sqrt{3}/\sqrt{3} to get 633=23\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20184 marksExpress 53−2\dfrac{5}{3 - \sqrt{2}} in the form a+b2a + b\sqrt{2}, where aa and bb are rational numbers.
Show worked answer →

Multiply numerator and denominator by the conjugate 3+23 + \sqrt{2} (M1).

Denominator: (3−2)(3+2)=9−2=7(3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7 (A1).

Numerator: 5(3+2)=15+525(3 + \sqrt{2}) = 15 + 5\sqrt{2} (M1).

So 53−2=15+527=157+572\dfrac{5}{3 - \sqrt{2}} = \dfrac{15 + 5\sqrt{2}}{7} = \dfrac{15}{7} + \dfrac{5}{7}\sqrt{2}, giving a=157a = \dfrac{15}{7} and b=57b = \dfrac{5}{7} (A1).

Markers reward multiplying by the conjugate, the rationalised denominator, the expanded numerator, and the final form with aa and bb identified.

OCR 20223 marksSimplify (8x627)−23\left(\dfrac{8x^6}{27}\right)^{-\frac{2}{3}}, giving your answer in the form pqxr\dfrac{p}{q x^r}.
Show worked answer →

A negative power means reciprocal, so (8x627)−23=(278x6)23\left(\dfrac{8x^6}{27}\right)^{-\frac{2}{3}} = \left(\dfrac{27}{8x^6}\right)^{\frac{2}{3}} (M1).

Apply the power to each part: 272/3=(271/3)2=32=927^{2/3} = (27^{1/3})^2 = 3^2 = 9, 82/3=(81/3)2=22=48^{2/3} = (8^{1/3})^2 = 2^2 = 4, and (x6)2/3=x4(x^6)^{2/3} = x^4 (M1).

So the result is 94x4\dfrac{9}{4x^4}, giving p=9p = 9, q=4q = 4, r=4r = 4 (A1).

Markers reward turning the negative power into a reciprocal, applying the cube-root-then-square correctly to 88 and 2727, and the final simplified fraction.

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