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Indices and surds

Laws of indices for rational exponents; simplifying surds and rationalising denominators.

Spec MM1.1, MM1.2, M2.7, M2.11, M2.12Section 1 Part 1: AS-level pure mathematics23 questions3 real past questions
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1Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Simplify \(\dfrac{(2x^3)^4 \times 8x^{-5}}{4x^2}\).

  1. A\(32x^{9}\)
  2. B\(16x^{7}\)
  3. C\(x^{5}\)
  4. D\(16x^{5}\)
  5. E\(4x^{5}\)
  6. F\(32x^{5}\)
Worked solution

\((2x^3)^4 = 16x^{12}\), so the numerator is \(16x^{12}\times 8x^{-5} = 128x^{7}\). Dividing by \(4x^2\) gives \(32x^{5}\).

Full method — how and why
A pure index-laws question. Nothing here needs thought once you treat the numbers and the powers of \(x\) as two separate bookkeeping jobs.
  1. Deal with the bracket first: \((2x^3)^4 = 2^4 x^{12} = 16x^{12}\).
    A power outside a bracket multiplies every factor inside, so the \(2\) is raised to the 4th power as well as the \(x^3\). This is the step people skip.
  2. Multiply the numerator: \(16x^{12} \times 8x^{-5} = 128x^{7}\).
    Multiplying powers of the same base adds the indices: \(12 + (-5) = 7\). The numbers just multiply normally.
  3. Divide by \(4x^2\): \(\dfrac{128x^{7}}{4x^{2}} = 32x^{5}\).
    Dividing subtracts indices: \(7 - 2 = 5\), and \(128 \div 4 = 32\).
Where it goes wrong
  • Forgetting to raise the \(2\) to the 4th power gives \(2x^{12}\) at step 1 and the answer \(4x^5\).
  • Subtracting instead of adding the \(-5\) gives \(x^{9}\) and the answer \(32x^{9}\) - which is offered.
Takeaway Handle the number part and the index part separately, and always distribute an outside power over everything in the bracket.
Answer F · Indices and surds (MM1.1) · question IND-01
2Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Find the value of \(\left(\dfrac{27}{8}\right)^{-\frac{2}{3}}\).

  1. A\(\dfrac{9}{4}\)
  2. B\(-\dfrac{4}{9}\)
  3. C\(\dfrac{4}{9}\)
  4. D\(-\dfrac{9}{4}\)
  5. E\(\dfrac{2}{3}\)
  6. F\(\dfrac{3}{2}\)
Worked solution

The negative index inverts the fraction: \(\left(\frac{8}{27}\right)^{2/3}\). Cube root first: \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

Full method — how and why
A fractional negative index of a fraction. Three separate instructions are packed into \(-\frac23\): invert, take the cube root, square.
  1. The minus sign inverts the fraction: \(\left(\frac{27}{8}\right)^{-2/3} = \left(\frac{8}{27}\right)^{2/3}\).
    \(a^{-n} = \frac{1}{a^n}\), and the reciprocal of a fraction is the fraction turned upside down. Doing this first keeps the numbers small.
  2. Take the cube root of top and bottom: \(\left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}\).
    \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\); rooting before powering keeps you working with \(2\) and \(3\) rather than \(64\) and \(729\).
  3. Square the result: \(\left(\frac23\right)^2 = \frac49\).
    The numerator \(2\) in the index is the power.
Where it goes wrong
  • Treating the minus sign as making the answer negative gives \(-\frac94\) or \(-\frac49\); a negative index never makes a positive number negative.
  • Forgetting to square gives \(\frac23\); forgetting to invert gives \(\frac94\). All three are offered as options.
Takeaway Read a fractional index as three instructions - sign (invert), denominator (root), numerator (power) - and do them in that order.
Answer C · Indices and surds (MM1.1) · question IND-02
3Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(4^{x+1} = 8^{2x-3}\).

  1. A\(x = \dfrac{11}{4}\)
  2. B\(x = \dfrac{5}{4}\)
  3. C\(x = \dfrac{9}{4}\)
  4. D\(x = \dfrac{7}{2}\)
  5. E\(x = 2\)
  6. F\(x = 4\)
Worked solution

Write both sides as powers of 2: \(2^{2x+2} = 2^{6x-9}\). So \(2x + 2 = 6x - 9\), giving \(4x = 11\) and \(x = \frac{11}{4}\).

Full method — how and why
An exponential equation where both sides can be written as powers of the same base. Once the bases match, the indices must be equal.
  1. Spot that \(4 = 2^2\) and \(8 = 2^3\).
    Both numbers are powers of \(2\), so the whole equation can be rewritten in base \(2\) and the exponentials compared directly.
  2. Rewrite: \(4^{x+1} = 2^{2(x+1)} = 2^{2x+2}\) and \(8^{2x-3} = 2^{3(2x-3)} = 2^{6x-9}\).
    \((a^m)^n = a^{mn}\): the outer index multiplies the whole inner index, so brackets are essential.
  3. Equate the indices: \(2x + 2 = 6x - 9\), so \(4x = 11\) and \(x = \frac{11}{4}\).
    \(2^u = 2^v\) forces \(u = v\) because \(2^x\) is a one-to-one function (it never takes the same value twice).
Where it goes wrong
  • Multiplying out as \(3 \times 2x - 3\) instead of \(3(2x - 3)\) loses the bracket and gives a different linear equation.
  • Writing \(8 = 2^4\) (a slip) gives \(x = \frac{7}{2}\), which is offered.
Takeaway Whenever an equation has different bases, look for a common base first; comparing indices is then legitimate because exponentials are one-to-one.
Answer A · Indices and surds (MM1.1) · question IND-03
4Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

Express \(\dfrac{5}{3+2\sqrt{5}}\) in the form \(a + b\sqrt{5}\) where \(a\) and \(b\) are rational.

  1. A\(15 - 10\sqrt{5}\)
  2. B\(\dfrac{-15 + 10\sqrt{5}}{11}\)
  3. C\(\dfrac{15 + 10\sqrt{5}}{29}\)
  4. D\(\dfrac{15 - 10\sqrt{5}}{11}\)
  5. E\(-15 + 10\sqrt{5}\)
  6. F\(\dfrac{-15 + 10\sqrt{5}}{29}\)
Worked solution

Multiply top and bottom by \(3 - 2\sqrt{5}\): denominator \(9 - 20 = -11\), numerator \(15 - 10\sqrt{5}\). So the value is \(\dfrac{15-10\sqrt{5}}{-11} = \dfrac{-15+10\sqrt{5}}{11}\).

Full method — how and why
Rationalising a denominator of the form \(p + q\sqrt{n}\). The tool is the difference of two squares, which turns the surd denominator into an integer.
  1. Multiply numerator and denominator by the conjugate \(3 - 2\sqrt5\).
    \((p + q\sqrt n)(p - q\sqrt n) = p^2 - q^2 n\), which contains no surd. Multiplying top and bottom by the same thing does not change the value.
  2. Denominator: \(3^2 - (2\sqrt5)^2 = 9 - 4 \times 5 = -11\).
    Square the whole second term, coefficient included: \((2\sqrt5)^2 = 4 \times 5\), not \(2 \times 5\).
  3. Numerator: \(5(3 - 2\sqrt5) = 15 - 10\sqrt5\).
    Nothing subtle - just expand.
  4. Tidy the signs: \(\dfrac{15 - 10\sqrt5}{-11} = \dfrac{-15 + 10\sqrt5}{11}\).
    A negative denominator is not 'in the form \(a + b\sqrt5\)' with the usual convention, and both versions are offered, so the sign handling decides the mark.
Where it goes wrong
  • \((2\sqrt5)^2 = 10\) (forgetting to square the \(2\)) gives denominator \(-1\) and a wildly different answer.
  • Leaving \(\frac{15 - 10\sqrt5}{-11}\) and picking the option with \(+11\) in the denominator but the wrong signs on top.
Takeaway Multiply by the conjugate, square the coefficient as well as the surd, then fix the sign of the denominator at the end.
Answer B · Indices and surds (MM1.2) · question IND-04
5Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

Simplify \(\sqrt{75} - \sqrt{27} + \sqrt{12}\).

  1. A\(2\sqrt{3}\)
  2. B\(6\sqrt{3}\)
  3. C\(\sqrt{60}\)
  4. D\(10\sqrt{3}\)
  5. E\(4\sqrt{3}\)
  6. F\(5\sqrt{3}\)
Worked solution

\(\sqrt{75} = 5\sqrt{3}\), \(\sqrt{27} = 3\sqrt{3}\), \(\sqrt{12} = 2\sqrt{3}\), so the total is \((5 - 3 + 2)\sqrt{3} = 4\sqrt{3}\).

Full method — how and why
Simplifying a sum of surds. They can only be combined once each is written as a multiple of the same surd.
  1. Pull out the largest square factor from each: \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt3\), \(\sqrt{27} = \sqrt{9\times3} = 3\sqrt3\), \(\sqrt{12} = \sqrt{4\times3} = 2\sqrt3\).
    \(\sqrt{ab} = \sqrt a\sqrt b\), so a perfect-square factor comes out as a whole number. All three reduce to multiples of \(\sqrt3\), which is why the sum simplifies at all.
  2. Collect like terms: \(5\sqrt3 - 3\sqrt3 + 2\sqrt3 = 4\sqrt3\).
    Once every term has the same surd part, \(\sqrt3\) behaves exactly like an algebraic \(x\).
Where it goes wrong
  • Adding under one root sign: \(\sqrt{75} - \sqrt{27} \neq \sqrt{48}\). Roots do not distribute over \(+\) or \(-\).
  • Using \(\sqrt{75} = 3\sqrt5\) (wrong square factor) - always take out the largest square.
Takeaway Surds add only when their irrational parts match, so simplify each one fully before trying to combine.
Answer E · Indices and surds (MM1.2) · question IND-05
6Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(9^{x} - 4\times 3^{x} + 3 = 0\).

  1. A\(x = 1\) only
  2. B\(x = 0\) or \(x = 1\)
  3. C\(x = -1\) or \(x = 0\)
  4. D\(x = 0\) only
  5. E\(x = \log_3 4\)
  6. F\(x = 1\) or \(x = 3\)
Worked solution

Let \(y = 3^x\): \(y^2 - 4y + 3 = (y-1)(y-3) = 0\), so \(3^x = 1\) or \(3^x = 3\), giving \(x = 0\) or \(x = 1\).

Full method — how and why
A hidden quadratic. The key observation is \(9^x = (3^2)^x = (3^x)^2\), so the equation is quadratic in \(3^x\).
  1. Write \(9^x = (3^x)^2\) and substitute \(y = 3^x\): \(y^2 - 4y + 3 = 0\).
    Spotting that one exponential is the square of the other is the whole question; without it there is no route in.
  2. Factorise: \((y - 1)(y - 3) = 0\), so \(y = 1\) or \(y = 3\).
    Ordinary quadratic technique once the substitution is made.
  3. Undo the substitution: \(3^x = 1 \Rightarrow x = 0\); \(3^x = 3 \Rightarrow x = 1\).
    \(a^0 = 1\) for any \(a\) - the root \(y = 1\) is easy to lose because it looks trivial.
Where it goes wrong
  • Stopping at \(y = 1, 3\) and offering those as the answers - the question asks for \(x\), not \(3^x\).
  • Rejecting \(y = 1\) as 'not a real solution'. Only non-positive values of \(y\) must be rejected, because \(3^x > 0\) always.
Takeaway If an equation contains \(a^{2x}\) and \(a^x\), substitute \(y = a^x\); afterwards discard only the roots with \(y \leq 0\).
Answer B · Indices and surds (MM1.1) · question IND-08
7Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

The value of \(\dfrac{2^{n+2} + 2^{n}}{2^{n+1}}\) is

  1. A\(2^{n}\)
  2. B\(5\)
  3. C\(\dfrac{3}{2}\)
  4. D\(3\)
  5. E\(\dfrac{5}{2}\)
  6. F\(2\)
Worked solution

Factor \(2^n\): numerator \(= 2^n(4 + 1) = 5\cdot 2^n\); denominator \(= 2\cdot 2^n\). Quotient \(= \frac52\).

Full method — how and why
Powers with an unknown exponent. Factorising out the smallest power turns the expression into a number.
  1. Write every term with the common factor \(2^n\): \(2^{n+2} = 4\cdot 2^n\), \(2^{n+1} = 2\cdot 2^n\).
    \(2^{n+k} = 2^k \cdot 2^n\). Choosing the *smallest* power (\(2^n\)) as the common factor makes every remaining factor a whole number.
  2. Substitute: \(\dfrac{4\cdot2^n + 2^n}{2\cdot 2^n} = \dfrac{5 \cdot 2^n}{2\cdot 2^n}\).
    Factorising the numerator is legitimate because both terms contain \(2^n\).
  3. Cancel \(2^n\): the value is \(\frac52\), independent of \(n\).
    \(2^n \neq 0\) for every \(n\), so cancelling is always valid here.
Where it goes wrong
  • 'Cancelling' the \(n\)'s in the indices, e.g. claiming \(\frac{2^{n+2}}{2^{n+1}} = 2^{2}/2^{1}\) term by term across a sum - you cannot split a fraction over a sum in the denominator.
  • Answering \(2^n\) because the \(n\) 'has to appear somewhere'.
Takeaway With unknown exponents, factor out the lowest power; the \(n\) almost always cancels and the answer is a plain number.
Answer E · Indices and surds (MM1.1) · question IND-10
8Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Simplify \(\left(\dfrac{a^{\frac{1}{2}} b^{-2}}{a^{-1} b^{\frac{1}{2}}}\right)^{4}\).

  1. A\(a^{2}b^{-10}\)
  2. B\(a^{2}b^{-6}\)
  3. C\(a^{6}b^{10}\)
  4. D\(a^{6}b^{-6}\)
  5. E\(a^{6}b^{-10}\)
  6. F\(a^{-2}b^{10}\)
Worked solution

Inside the bracket: \(a^{\frac12 + 1} b^{-2 - \frac12} = a^{\frac32} b^{-\frac52}\). Raise to the 4th power: \(a^{6} b^{-10}\).

Full method — how and why
Index laws with two variables. Simplify inside the bracket first, then apply the outside power to each index.
  1. Simplify the \(a\)'s inside: \(\dfrac{a^{1/2}}{a^{-1}} = a^{1/2 - (-1)} = a^{3/2}\).
    Dividing subtracts indices, and subtracting a negative adds - the commonest slip in the question.
  2. Simplify the \(b\)'s: \(\dfrac{b^{-2}}{b^{1/2}} = b^{-2 - 1/2} = b^{-5/2}\).
    Same rule; keep the two variables in separate columns so signs do not get mixed up.
  3. Raise the bracket to the 4th power: \(\left(a^{3/2}b^{-5/2}\right)^4 = a^{6}b^{-10}\).
    \((a^m)^n = a^{mn}\) applies to each base separately: \(\frac32 \times 4 = 6\) and \(-\frac52 \times 4 = -10\).
Where it goes wrong
  • Getting \(a^{-1/2}\) at step 1 by subtracting the wrong way round, leading to \(a^{-2}b^{10}\) - which is offered.
  • Applying the power \(4\) to only the first factor.
Takeaway Simplify inside the bracket to a single power of each base before applying the outer index.
Answer E · Indices and surds (MM1.1) · question IND-13
9Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

The number \(\dfrac{\sqrt{50} + \sqrt{18}}{\sqrt{8}}\) is equal to

  1. A\(4\)
  2. B\(\dfrac{8}{\sqrt 2}\)
  3. C\(\sqrt{34}\)
  4. D\(8\)
  5. E\(2\sqrt{2}\)
  6. F\(4\sqrt{2}\)
Worked solution

\(\sqrt{50} = 5\sqrt2\), \(\sqrt{18} = 3\sqrt2\), \(\sqrt8 = 2\sqrt2\). So \(\dfrac{8\sqrt2}{2\sqrt2} = 4\).

Full method — how and why
A quotient of surds that collapses completely. Simplify each root first and the \(\sqrt2\)'s cancel.
  1. Simplify: \(\sqrt{50} = 5\sqrt2\), \(\sqrt{18} = 3\sqrt2\), \(\sqrt8 = 2\sqrt2\).
    Every number is (square) \(\times\, 2\), so every root is a multiple of \(\sqrt2\) - which is why the fraction will be rational.
  2. Add the numerator: \(5\sqrt2 + 3\sqrt2 = 8\sqrt2\).
    Like terms again - \(\sqrt2\) behaves as a common factor.
  3. Divide: \(\dfrac{8\sqrt2}{2\sqrt2} = 4\).
    The \(\sqrt2\) cancels top and bottom, leaving plain arithmetic.
Where it goes wrong
  • Working out \(\sqrt{50} + \sqrt{18} = \sqrt{68}\), which is wrong and leads nowhere.
  • Cancelling only the numbers and leaving a stray \(\sqrt2\), giving \(4\sqrt2\).
Takeaway Simplify every surd to the form \(k\sqrt m\) before adding or dividing; common surd factors then cancel visibly.
Answer A · Indices and surds (MM1.2) · question IND-14
10Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Given that \(2^{x} = 5\) and \(2^{y} = 20\), find the value of \(2^{2x - y}\).

  1. A\(\dfrac{1}{4}\)
  2. B\(\dfrac{4}{5}\)
  3. C\(\dfrac{5}{4}\)
  4. D\(-10\)
  5. E\(\dfrac{25}{2}\)
  6. F\(5\)
Worked solution

\(2^{2x - y} = \dfrac{(2^x)^2}{2^y} = \dfrac{25}{20} = \dfrac54\).

Full method — how and why
You are given two exponentials and asked for a third. Express the target in terms of the given ones using index laws - you never need to find \(x\) or \(y\).
  1. Split the index: \(2^{2x - y} = \dfrac{2^{2x}}{2^{y}}\).
    Subtraction in the index is division - this is what lets you use the two given facts separately.
  2. Write \(2^{2x} = (2^x)^2 = 5^2 = 25\).
    A coefficient in the index is a power of the whole exponential: \(2^{2x} = (2^x)^2\), not \(2 \times 2^x\).
  3. Substitute: \(\dfrac{25}{20} = \dfrac54\).
    \(2^y = 20\) is given directly.
Where it goes wrong
  • Using \(2^{2x} = 2 \times 2^x = 10\), giving \(\frac12\).
  • Trying to solve for \(x = \log_2 5\) and \(y = \log_2 20\) first - correct but slow, and it invites arithmetic slips.
Takeaway Rewrite the target index as sums and multiples of the given ones; substitution then beats solving for the unknowns.
Answer C · Indices and surds (MM1.1) · question IND-15
11Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(x^{\frac{3}{2}} = 8x^{-\frac{3}{2}}\) for \(x > 0\).

  1. A\(x = 2\sqrt{2}\)
  2. B\(x = 16\)
  3. C\(x = 8\)
  4. D\(x = \sqrt{2}\)
  5. E\(x = 2\)
  6. F\(x = 4\)
Worked solution

Multiply by \(x^{3/2}\): \(x^3 = 8\), so \(x = 2\).

Full method — how and why
An equation with the unknown in two fractional powers. Multiplying by the negative power clears it entirely.
  1. Multiply both sides by \(x^{3/2}\): \(x^{3/2} \cdot x^{3/2} = 8\), i.e. \(x^3 = 8\).
    \(x^{-3/2}\) becomes \(x^0 = 1\) on the right, and the left gains \(\frac32 + \frac32 = 3\). Multiplying by a positive quantity (we are told \(x > 0\)) cannot introduce false solutions.
  2. Take the cube root: \(x = 2\).
    \(x^3 = 8\) has one real root.
Where it goes wrong
  • Cancelling the powers to get \(x = 8\).
  • Dividing by \(x^{3/2}\) instead of multiplying, which leaves \(x^{3} = 8\) on the wrong side and invites sign errors in the index.
Takeaway To clear a negative fractional power, multiply through by that power - stated positivity guarantees the step is reversible.
Answer E · Indices and surds (MM1.1) · question IND-17
12Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

How many real solutions does the equation \(2^{x^2} = 16 \times 2^{-3x}\) have?

  1. A\(3\)
  2. B\(2\)
  3. C\(4\)
  4. D\(1\)
  5. E\(0\)
  6. Finfinitely many
Worked solution

\(2^{x^2} = 2^{4 - 3x}\), so \(x^2 + 3x - 4 = 0\), i.e. \((x+4)(x-1) = 0\): two real solutions.

Full method — how and why
'How many real solutions' with exponentials - convert to a single base and the exponents give a quadratic whose discriminant answers the question.
  1. Write \(16 = 2^4\), so the right-hand side is \(2^{4}\cdot2^{-3x} = 2^{4 - 3x}\).
    Multiplying powers of the same base adds indices; this puts both sides in base \(2\).
  2. Equate indices: \(x^2 = 4 - 3x\), i.e. \(x^2 + 3x - 4 = 0\).
    Legitimate because \(2^u\) is one-to-one.
  3. Factorise: \((x+4)(x-1) = 0\) - two distinct real roots.
    The question asks only for the *number* of solutions, so the discriminant (\(9 + 16 > 0\)) would do; factorising confirms both roots are real.
Where it goes wrong
  • Rejecting \(x = -4\) because it is negative - the exponent may be negative; it is \(2^x\) that must stay positive.
  • Reading \(16 \times 2^{-3x}\) as \((16 \times 2)^{-3x}\).
Takeaway Same base \(\Rightarrow\) equate exponents; then it is an ordinary quadratic and every real root counts.
Answer B · Indices and surds (MM1.1) · question IND-19
13Indices and surds · spec MM1.2 · Paper 1 style · ★☆☆

Given that \(\sqrt{a} - \sqrt{b} = 2\) and \(a - b = 12\), find \(\sqrt{a} + \sqrt{b}\).

  1. A\(4\)
  2. B\(8\)
  3. C\(10\)
  4. D\(6\)
  5. E\(24\)
  6. F\(3\)
Worked solution

\(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\), so \(12 = 2(\sqrt a + \sqrt b)\), giving \(6\).

Full method — how and why
You are given a difference of surds and a difference of the numbers. The link is the difference of two squares - no need to find \(a\) and \(b\).
  1. Factorise \(a - b\) as a difference of two squares in \(\sqrt a\) and \(\sqrt b\): \(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\).
    \(a = (\sqrt a)^2\) for \(a \geq 0\), so the whole expression is \(X^2 - Y^2\) with \(X = \sqrt a\), \(Y = \sqrt b\) - this is the only structural idea in the question.
  2. Substitute the given values: \(12 = 2 \times (\sqrt a + \sqrt b)\).
    The first bracket is exactly the quantity given as \(2\).
  3. Divide: \(\sqrt a + \sqrt b = 6\).
    Direct.
Where it goes wrong
  • Solving simultaneously for \(a\) and \(b\) (possible: \(a = 16\), \(b = 4\)) - correct but three times the work and easy to fumble.
  • Answering \(24\) by multiplying rather than dividing.
Takeaway \(a - b = (\sqrt a - \sqrt b)(\sqrt a + \sqrt b)\) converts between sums and differences of surds instantly.
Answer D · Indices and surds (MM1.2) · question IND-20
14Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

If \(x = 2^{p}\), \(y = 2^{q}\) and \(\dfrac{x^{3}}{y^{2}} = 2^{5}\) with \(xy = 2^{10}\), find \(p\).

  1. A\(3\)
  2. B\(7\)
  3. C\(8\)
  4. D\(4\)
  5. E\(6\)
  6. F\(5\)
Worked solution

\(3p - 2q = 5\) and \(p + q = 10\). From the second, \(q = 10 - p\); then \(3p - 20 + 2p = 5\), so \(5p = 25\) and \(p = 5\).

Full method — how and why
Two equations in disguise. Writing both conditions in base \(2\) turns them into simultaneous linear equations in \(p\) and \(q\).
  1. \(\dfrac{x^3}{y^2} = \dfrac{2^{3p}}{2^{2q}} = 2^{3p - 2q}\), so \(3p - 2q = 5\).
    Powers of powers multiply indices, division subtracts them; equating indices is valid because base \(2\) is one-to-one.
  2. \(xy = 2^{p}\cdot 2^{q} = 2^{p+q}\), so \(p + q = 10\).
    Multiplication adds indices.
  3. Solve: from the second, \(q = 10 - p\); substituting, \(3p - 2(10 - p) = 5 \Rightarrow 5p = 25 \Rightarrow p = 5\).
    Substitution is quickest with only two unknowns; note the bracket when substituting a negative.
Where it goes wrong
  • Dropping the bracket to get \(3p - 20 - 2p = 5\) and hence \(p = 25\).
  • Equating \(x^3/y^2 = 2^5\) to \(3p/2q = 5\) - indices subtract, they do not divide.
Takeaway Converting everything to one base turns exponential conditions into linear equations in the exponents.
Answer F · Indices and surds (MM1.1) · question IND-21
15Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Given that \((2 + \sqrt{3})^2 = a + b\sqrt{3}\) and \((2+\sqrt{3})^{-1} = c + d\sqrt{3}\), where \(a\), \(b\), \(c\), \(d\) are integers, find the value of \(a + b + c + d\).

  1. A\(14\)
  2. B\(4\)
  3. C\(6\)
  4. D\(10\)
  5. E\(12\)
  6. F\(8\)
Worked solution

\((2+\sqrt3)^2 = 7 + 4\sqrt3\), so \(a = 7\), \(b = 4\). \((2+\sqrt3)^{-1} = \dfrac{2-\sqrt3}{4-3} = 2 - \sqrt3\), so \(c = 2\), \(d = -1\). Sum \(= 7 + 4 + 2 - 1 = 12\).

Full method — how and why
Two standard surd manipulations in one: squaring a bracket and rationalising a reciprocal. The question then just asks you to keep track of the four integers.
  1. Square the bracket: \((2 + \sqrt3)^2 = 4 + 4\sqrt3 + 3 = 7 + 4\sqrt3\), so \(a = 7\), \(b = 4\).
    Use \((x+y)^2 = x^2 + 2xy + y^2\) with \(y = \sqrt3\), and remember \((\sqrt3)^2 = 3\) - that is what makes the rational part \(4 + 3\).
  2. Rationalise the reciprocal: \(\dfrac{1}{2+\sqrt3} \times \dfrac{2 - \sqrt3}{2-\sqrt3} = \dfrac{2-\sqrt3}{4 - 3} = 2 - \sqrt3\), so \(c = 2\), \(d = -1\).
    The conjugate gives denominator \(2^2 - 3 = 1\), which is why this particular reciprocal is so clean.
  3. Add: \(7 + 4 + 2 + (-1) = 12\).
    The question wants the sum, so the sign of \(d\) matters.
Where it goes wrong
  • Taking \(d = 1\) and getting \(14\) - the rationalised form is \(2 - \sqrt3\), so \(d\) is negative.
  • Writing \((2+\sqrt3)^2 = 4 + 3 = 7\) (forgetting the cross term \(2 \times 2 \times \sqrt3\)).
Takeaway \((a+\sqrt b)\) and \((a - \sqrt b)\) multiply to the integer \(a^2 - b\); that single fact drives both halves of this question.
Answer E · Indices and surds (MM1.2) · question IND-06
16Indices and surds · spec MM1.1 · Paper 1 style · ★★☆

Which of the following is equal to \(\dfrac{x^2 \sqrt{x}}{\sqrt[3]{x}}\) for \(x > 0\)?

  1. A\(x^{3}\)
  2. B\(x^{\frac{5}{2}}\)
  3. C\(x^{\frac{7}{3}}\)
  4. D\(x^{\frac{11}{6}}\)
  5. E\(x^{\frac{17}{6}}\)
  6. F\(x^{\frac{13}{6}}\)
Worked solution

The index is \(2 + \frac12 - \frac13 = \frac{12 + 3 - 2}{6} = \frac{13}{6}\).

Full method — how and why
Roots disguised as indices. Convert every root to a fractional power and the question becomes adding and subtracting fractions.
  1. Rewrite the roots: \(\sqrt x = x^{1/2}\) and \(\sqrt[3]{x} = x^{1/3}\).
    \(\sqrt[n]{x} = x^{1/n}\). In index form the laws apply; in root form they do not.
  2. Combine: \(\dfrac{x^2 \cdot x^{1/2}}{x^{1/3}} = x^{2 + \frac12 - \frac13}\).
    Multiplication adds indices, division subtracts them - the three operations become one sum.
  3. Add the fractions over a common denominator \(6\): \(2 + \frac12 - \frac13 = \frac{12 + 3 - 2}{6} = \frac{13}{6}\).
    \(6\) is the lowest common denominator of \(1\), \(2\) and \(3\); converting \(2\) to \(\frac{12}{6}\) is where slips happen.
Where it goes wrong
  • Adding the \(\frac13\) instead of subtracting gives \(x^{17/6}\), which is offered.
  • Forgetting to convert the whole number \(2\) into sixths.
Takeaway Turn every root into a fractional index immediately; then all you are doing is arithmetic with fractions.
Answer F · Indices and surds (MM1.1) · question IND-07
17Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Find the value of \(\dfrac{1}{\sqrt{2}+1} + \dfrac{1}{\sqrt{3}+\sqrt{2}} + \dfrac{1}{2+\sqrt{3}}\).

  1. A\(2\)
  2. B\(2 - \sqrt{3}\)
  3. C\(\sqrt{3} + 1\)
  4. D\(1\)
  5. E\(\sqrt{3}\)
  6. F\(\sqrt{2}\)
Worked solution

Rationalising each term gives \((\sqrt2 - 1) + (\sqrt3 - \sqrt2) + (2 - \sqrt3)\), which telescopes to \(1\).

Full method — how and why
A telescoping sum. Each term rationalises to a difference of two surds, and almost everything cancels - so do not compute decimals.
  1. Rationalise each term: \(\dfrac{1}{\sqrt2+1} = \sqrt2 - 1\), \(\dfrac{1}{\sqrt3+\sqrt2} = \sqrt3 - \sqrt2\), \(\dfrac{1}{2+\sqrt3} = 2 - \sqrt3\).
    Each denominator has the form \(\sqrt{n+1} + \sqrt n\) whose conjugate gives \((n+1) - n = 1\) - the denominators all become \(1\), which is the point of the construction.
  2. Add them: \((\sqrt2 - 1) + (\sqrt3 - \sqrt2) + (2 - \sqrt3)\).
    Write the terms in a line so the cancelling pairs are visible.
  3. Cancel: \(\sqrt2\) with \(-\sqrt2\), \(\sqrt3\) with \(-\sqrt3\), leaving \(-1 + 2 = 1\).
    This is a telescoping sum - the surds annihilate in pairs and only the two ends survive.
Where it goes wrong
  • Approximating numerically and choosing \(\sqrt2\) or \(\sqrt3\) because the total 'looks about right'.
  • Rationalising with the wrong conjugate sign, which flips one term and destroys the cancellation.
Takeaway \(\dfrac{1}{\sqrt{n+1}+\sqrt n} = \sqrt{n+1} - \sqrt n\): recognise this and long chains of such fractions collapse instantly.
Answer D · Indices and surds (MM1.2) · question IND-09
18Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Given that \(x = \sqrt{7} + \sqrt{3}\) and \(y = \sqrt{7} - \sqrt{3}\), find \(x^2 + y^2 - xy\).

  1. A\(24\)
  2. B\(16\)
  3. C\(4\)
  4. D\(20\)
  5. E\(20 - 2\sqrt{21}\)
  6. F\(12\)
Worked solution

\(x^2 + y^2 = (10 + 2\sqrt{21}) + (10 - 2\sqrt{21}) = 20\) and \(xy = 7 - 3 = 4\). So \(20 - 4 = 16\).

Full method — how and why
A symmetric surd expression. Computing \(x^2\), \(y^2\) and \(xy\) separately is much cleaner than expanding everything, because the irrational parts cancel.
  1. \(x^2 = (\sqrt7+\sqrt3)^2 = 7 + 2\sqrt{21} + 3 = 10 + 2\sqrt{21}\) and \(y^2 = 10 - 2\sqrt{21}\).
    The cross terms have opposite signs because \(y\) is the conjugate of \(x\) - that is what makes the sum rational.
  2. Add: \(x^2 + y^2 = 20\).
    The \(\pm 2\sqrt{21}\) terms cancel exactly.
  3. \(xy = (\sqrt7+\sqrt3)(\sqrt7-\sqrt3) = 7 - 3 = 4\).
    Difference of two squares again - no cross terms at all.
  4. Combine: \(x^2 + y^2 - xy = 20 - 4 = 16\).
    Subtract in the order the question asks.
Where it goes wrong
  • Computing \(x^2 + y^2\) as \((x+y)^2\) and getting \(28\) - remember \((x+y)^2 = x^2 + y^2 + 2xy\).
  • Giving \(20 - 2\sqrt{21}\) by forgetting that the surds cancel when the squares are added.
Takeaway With conjugate pairs, \(x + y\), \(xy\) and \(x^2 + y^2\) are all rational; build the expression from those rather than expanding blindly.
Answer B · Indices and surds (MM1.2) · question IND-11
19Indices and surds · spec MM1.2 · Paper 2 style · ★★☆

Which of the following statements is/are true for all positive real numbers \(a\) and \(b\)?

  1. I\(\sqrt{a} \sqrt{b} = \sqrt{ab}\)
  2. II\(\sqrt{a + b} = \sqrt{a} + \sqrt{b}\)
  3. III\(\sqrt{a^2 + b^2} < a + b\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is a law of surds. II fails for \(a = b = 1\): \(\sqrt2 \neq 2\). III: \((a+b)^2 = a^2 + b^2 + 2ab > a^2 + b^2\) since \(ab > 0\), and both sides are positive, so III is true.

Full method — how and why
Three surd 'rules', two of which are genuine. The exam skill is knowing which laws distribute and finding a counterexample for the one that does not.
  1. Statement I: \(\sqrt a\sqrt b = \sqrt{ab}\) is a law of surds, valid for all \(a, b \geq 0\). True.
    Roots distribute over multiplication and division, and the condition \(a, b > 0\) in the question is exactly what the law needs.
  2. Statement II: test \(a = b = 1\): \(\sqrt{1+1} = \sqrt2 \approx 1.41\) but \(\sqrt1 + \sqrt1 = 2\). False.
    One counterexample is enough to kill a 'for all' statement - and the simplest numbers usually work.
  3. Statement III: square both sides. \((a+b)^2 = a^2 + b^2 + 2ab > a^2 + b^2\) since \(ab > 0\), and both sides are positive, so \(\sqrt{a^2+b^2} < a + b\). True.
    Squaring is safe here precisely because both sides are known to be positive; that is what makes the comparison valid in both directions.
  4. I and III are true: the answer is 'I and III only'.
    Match your three verdicts to the option list rather than reasoning about the options themselves.
Where it goes wrong
  • Assuming roots distribute over addition because they distribute over multiplication - II is the classic error this question is built around.
  • Rejecting III because the triangle-inequality picture feels like it should be an equality.
Takeaway Roots split over \(\times\) and \(\div\) but never over \(+\) or \(-\); to compare positive quantities, square them.
Answer F · Indices and surds (MM1.2) · question IND-16
20Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

The expression \(\dfrac{3}{7 - 2\sqrt{3}} - \dfrac{3}{7 + 2\sqrt{3}}\) simplifies to

  1. A\(\dfrac{12\sqrt{3}}{61}\)
  2. B\(\dfrac{6\sqrt{3}}{37}\)
  3. C\(\dfrac{42}{61}\)
  4. D\(\dfrac{42}{37}\)
  5. E\(0\)
  6. F\(\dfrac{12\sqrt{3}}{37}\)
Worked solution

Common denominator \((7-2\sqrt3)(7+2\sqrt3) = 49 - 12 = 37\). Numerator: \(3(7 + 2\sqrt3) - 3(7 - 2\sqrt3) = 12\sqrt3\).

Full method — how and why
A difference of two rationalisable fractions. Using a single common denominator is faster than rationalising each one.
  1. Use the common denominator \((7 - 2\sqrt3)(7 + 2\sqrt3) = 49 - 12 = 37\).
    The two denominators are conjugates, so their product is automatically rational - no separate rationalising needed.
  2. Numerator: \(3(7 + 2\sqrt3) - 3(7 - 2\sqrt3) = 3 \times 4\sqrt3 = 12\sqrt3\).
    When you subtract conjugates the rational parts cancel and the surd parts double - so only the surd survives.
  3. Result: \(\dfrac{12\sqrt3}{37}\).
    Already rationalised, since the denominator is an integer.
Where it goes wrong
  • \((2\sqrt3)^2 = 4 \times 3 = 12\), not \(6\); using \(6\) gives denominator \(43\).
  • Adding the numerators instead of subtracting, giving \(\frac{42}{37}\) - which is offered.
Takeaway A difference of conjugate fractions leaves only the surd part; their product denominator is rational from the start.
Answer F · Indices and surds (MM1.2) · question IND-18
21Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Which one of the following is the largest?

  1. A\(2\sqrt{4.5}\)
  2. B\(\sqrt{2} + \sqrt{7}\)
  3. C\(\sqrt{3} + \sqrt{6}\)
  4. D\(\sqrt{4} + \sqrt{5}\)
  5. E\(\sqrt{1} + \sqrt{8}\)
Worked solution

Each option is \(\sqrt a + \sqrt b\) with \(a + b = 9\). Squaring gives \(9 + 2\sqrt{ab}\), which is largest when the product \(ab\) is largest, i.e. when \(a\) and \(b\) are as close as possible. Option E has \(a = b = 4.5\) (product \(20.25\)), beating \(4 \times 5 = 20\) for option C. Numerically: E \(= \sqrt{18} \approx 4.243\), C \(= 2 + \sqrt5 \approx 4.236\).

Full method — how and why
A comparison question. Every option has the form \(\sqrt a + \sqrt b\) with \(a + b = 9\) - spotting that shared structure turns guesswork into a one-line argument.
  1. Notice that in every option the two numbers under the roots add to \(9\) (including \(2\sqrt{4.5} = \sqrt{4.5} + \sqrt{4.5}\)).
    When all the candidates share a constraint, compare them through that constraint rather than by estimating each one.
  2. Square a general option: \((\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab} = 9 + 2\sqrt{ab}\).
    All the options are positive, so the largest option is the one with the largest square - and only the \(ab\) term varies.
  3. With a fixed sum, \(ab\) is largest when \(a\) and \(b\) are closest together. The closest pair here is \(a = b = 4.5\), giving \(ab = 20.25\), beating \(4 \times 5 = 20\).
    This is the standard 'fixed perimeter, maximum area' fact; it decides the question without any decimals.
  4. So \(2\sqrt{4.5} = \sqrt{18} \approx 4.243\) is the largest (option C, \(2 + \sqrt5 \approx 4.236\), is a close second).
    Worth a sanity check because the top two are so close.
Where it goes wrong
  • Estimating roots to one decimal place - \(4.24\) versus \(4.24\) does not separate the top two options.
  • Assuming \(\sqrt1 + \sqrt8\) is biggest because it contains the largest single number; the spread-out pair is actually the smallest.
Takeaway For a fixed sum \(a + b\), the product \(ab\) - and hence \(\sqrt a + \sqrt b\) - is maximised when \(a\) and \(b\) are as close as possible.
Answer A · Indices and surds (MM1.2) · question IND-22
22Indices and surds · spec MM1.1 · Paper 2 style · ★★☆

\(a\) and \(b\) are positive real numbers. Which of the following statements is/are true?

  1. IIf \(a^2 = b^2\) then \(a = b\).
  2. IIIf \(2^a = 2^b\) then \(a = b\).
  3. IIIIf \(a^b = 1\) then \(a = 1\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I holds because \(a, b > 0\) rules out \(a = -b\). II holds because \(2^x\) is one-to-one. III: for positive \(a\) and \(b\), \(a^b = 1\) gives \(b\log a = 0\), so \(\log a = 0\) and \(a = 1\). All three are true.

Full method — how and why
Three cancellation claims, all made safe by the stated positivity. Check what each would need.
  1. I: \(a^2=b^2\) gives \(a = \pm b\) in general - but \(a,b>0\) rules out \(a=-b\), so \(a=b\). True.
    The positivity hypothesis is exactly what repairs the usual counterexample.
  2. II: \(2^x\) is a strictly increasing function, so it takes each value once: \(2^a=2^b \Rightarrow a=b\). True.
    One-to-one functions can be cancelled; equivalently take \(\log_2\) of both sides.
  3. III: \(a^b = 1\) with \(a,b>0\). Take logs: \(b\log a = 0\). Since \(b>0\), \(\log a = 0\), so \(a=1\). True.
    Taking logs turns the exponent into a factor, and a product is zero only if a factor is.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting I by recalling the counterexample \(2\) and \(-2\) - forbidden here, since both are positive.
  • Rejecting III by thinking of \(a^0=1\) - that needs \(b=0\), which positivity forbids. (And \((-1)^2 = 1\) needs a negative \(a\).)
  • Overlooking how much work the phrase 'positive real numbers' is doing in all three.
Takeaway Read the domain restrictions first - positivity makes squaring and exponentiation reversible, when in general they are not.
Answer H · Indices and surds (MM1.1) · question X2-12
23Indices and surds · spec MM1.1 · Paper 2 style · ★★★

For how many integer values of \(n\) with \(1 \leq n \leq 100\) is \(\sqrt{n} + \sqrt{n+1}\) rational?

  1. A\(9\)
  2. B\(1\)
  3. C\(10\)
  4. D\(0\)
  5. E\(2\)
  6. F\(100\)
Worked solution

For the sum to be rational, both \(\sqrt n\) and \(\sqrt{n+1}\) would need to be rational (if one is irrational, \(\sqrt n + \sqrt{n+1}\) squared gives \(2n + 1 + 2\sqrt{n(n+1)}\) and \(n(n+1)\) is never a perfect square for \(n \geq 1\) since \(n^2 < n(n+1) < (n+1)^2\)). No two consecutive positive integers are both squares, so the answer is \(0\).

Full method — how and why
A 'for how many' question that is really about irrationality. The answer is \(0\), and seeing why needs one sharp observation about consecutive integers.
  1. Suppose \(\sqrt n + \sqrt{n+1} = r\) is rational and square it: \(2n + 1 + 2\sqrt{n(n+1)} = r^2\).
    Squaring is the standard way to get at a sum of surds; the rational parts separate from the single remaining surd.
  2. Rearrange: \(\sqrt{n(n+1)} = \dfrac{r^2 - 2n - 1}{2}\), so \(n(n+1)\) would have to be a perfect square.
    A rational square root of an integer is an integer, so \(n(n+1)\) must be a perfect square.
  3. But \(n^2 < n(n+1) < (n+1)^2\) for every \(n \geq 1\), so \(n(n+1)\) lies strictly between consecutive squares.
    There is no integer square strictly between \(n^2\) and \((n+1)^2\) - this is the contradiction.
  4. Hence no \(n\) in \(1 \leq n \leq 100\) works: the answer is \(0\).
    The bound \(100\) is a red herring; the argument rules out every positive integer.
Where it goes wrong
  • Testing \(n = 3\) (\(\sqrt3 + 2\)) and concluding 'sometimes it works' because one of the two roots is nice.
  • Answering \(9\) or \(10\) by counting perfect squares up to \(100\) - that would answer a different question.
Takeaway \(n(n+1)\) is never a perfect square for \(n \geq 1\), because it is squeezed strictly between \(n^2\) and \((n+1)^2\).
Answer D · Indices and surds (MM1.1) · question IND-12

How this topic has been examined

Real TMUA questions on Indices and surds, newest first. Links open the official paper at the question and the online marking page.

2022 · Paper 1Question 13surd expression with a and bquestion papersit online
2020 · Paper 1Question 7simultaneous equations in powers of 2question papersit online
2019 · Paper 2Question 19nested surdsquestion papersit online