Indices and surds - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 16 exam-style questions on Indices and surds for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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2–7 marks each
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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A student simplifies \(\left(8x^6\right)^{-\frac{2}{3}}\), where \(x > 0\), as follows.

\(\left(8x^6\right)^{-\frac{2}{3}} = 8^{-\frac{2}{3}} \times x^{6 - \frac{2}{3}} = -\dfrac{16}{3}x^{\frac{16}{3}}\)

Identify the two errors in the student's working.

Write \(\left(8x^6\right)^{-\frac{2}{3}}\) in the form \(ax^n\), where \(a\) and \(n\) are rational numbers.

Model answer

First error: \(\left(x^6\right)^{-\frac{2}{3}}\) is a power of a power, so the indices are multiplied: \(x^{6 \times (-\frac{2}{3})} = x^{-4}\). The student added them.
Second error: \(8^{-\frac{2}{3}}\) is not \(-\dfrac{2}{3} \times 8 = -\dfrac{16}{3}\). Correctly, \(8^{-\frac{2}{3}} = \dfrac{1}{8^{\frac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{8}\right)^2} = \dfrac{1}{2^2} = \dfrac{1}{4}\).
So \(\left(8x^6\right)^{-\frac{2}{3}} = \dfrac{1}{4}x^{-4}\).

Explanation

A power of a power means multiplying the indices, so \(\left(x^6\right)^{-\frac{2}{3}} = x^{-4}\), and \(8^{-\frac{2}{3}}\) means the reciprocal of the cube root of 8, squared, which is \(\dfrac{1}{4}\). The student made both of the most common index errors: adding instead of multiplying, and multiplying the base by the index. The correct simplification is \(\dfrac{1}{4}x^{-4}\).

Mark scheme

1 mark
B1 Identifies the error in the power of \(x\): \(\left(x^6\right)^{-\frac{2}{3}}\) needs the indices multiplied, \(6 \times \left(-\dfrac{2}{3}\right) = -4\), not added (the student found \(6 - \dfrac{2}{3}\))
The errors may be given in either order. Each B1 needs the error identified with a correct statement of what should have been done. "The student got the power wrong" alone scores B0.
1 mark
B1 Identifies the error in the number: \(8^{-\frac{2}{3}}\) is not \(-\dfrac{2}{3} \times 8\); it is \(\dfrac{1}{\left(\sqrt[3]{8}\right)^2} = \dfrac{1}{4}\) (a negative index gives a reciprocal, not a negative number)
1 mark
B1 \(\dfrac{1}{4}x^{-4}\)
Accept \(a = \dfrac{1}{4}\) and \(n = -4\), or \(0.25x^{-4}\). Accept \(\dfrac{1}{4x^4}\) only if \(a = \dfrac{1}{4}\) and \(n = -4\) are also stated.

Question 2

4 marks

Solutions relying entirely on calculator technology are not acceptable.

Given that \(3^p = 5\), find the exact value of each of
\[9^{p + 1} \qquad 27^{-\frac{2p}{3}} \qquad 3^{1 + \frac{p}{2}}\]

Model answer

\(9^{p + 1} = \left(3^2\right)^{p + 1} = 3^{2p + 2} = \left(3^p\right)^2 \times 3^2 = 25 \times 9 = 225\)
\(27^{-\frac{2p}{3}} = \left(3^3\right)^{-\frac{2p}{3}} = 3^{-2p} = \left(3^p\right)^{-2} = 5^{-2} = \dfrac{1}{25}\)
\(3^{1 + \frac{p}{2}} = 3 \times 3^{\frac{p}{2}} = 3 \times \left(3^p\right)^{\frac{1}{2}} = 3\sqrt{5}\)

Explanation

Write each number as a power of 3 and rearrange the index so that \(3^p\) appears: \(9^{p + 1} = \left(3^p\right)^2 \times 9\), \(27^{-\frac{2p}{3}} = \left(3^p\right)^{-2}\) and \(3^{1 + \frac{p}{2}} = 3\left(3^p\right)^{\frac{1}{2}}\). Then replace \(3^p\) by 5. A common error is to treat \(9^p\) as if it were also equal to 5.

Mark scheme

1 mark
M1 Writes \(9^{p + 1}\) in terms of \(3^p\), e.g. \(9^{p + 1} = 3^{2p + 2} = \left(3^p\right)^2 \times 3^2\) or \(9^{p + 1} = 9 \times \left(3^2\right)^p\). An answer with no working scores 0 marks.
An answer of 225, \(\dfrac{1}{25}\) and \(3\sqrt{5}\) with no working scores 0 marks: solutions relying entirely on calculator technology are not acceptable (finding \(p = \log_3 5 = 1.46\ldots\) and substituting is calculator work). Common error: \(9^{p + 1} = 9 \times 5 = 45\) (taking \(9^p\) to be 5).
1 mark
A1 225
1 mark
B1 \(27^{-\frac{2p}{3}} = 3^{-2p} = \left(3^p\right)^{-2} = \dfrac{1}{25}\)
Accept 0.04. A correct method must be seen.
1 mark
B1 \(3^{1 + \frac{p}{2}} = 3 \times \left(3^p\right)^{\frac{1}{2}} = 3\sqrt{5}\)
A correct method must be seen. Do not accept the decimal 6.71 only.

Question 3

4 marks

Using algebra, solve the equation
\[\dfrac{4^{x^2}}{2^x} = 8^{x + 2}\]

Model answer

\(4^{x^2} = \left(2^2\right)^{x^2} = 2^{2x^2}\) and \(8^{x + 2} = \left(2^3\right)^{x + 2} = 2^{3x + 6}\)
\(\dfrac{2^{2x^2}}{2^x} = 2^{2x^2 - x}\), so \(2^{2x^2 - x} = 2^{3x + 6}\)
\(2x^2 - x = 3x + 6\)
\(2x^2 - 4x - 6 = 0\), so \(x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0\)
\(x = 3\) or \(x = -1\)

Explanation

Write every term as a power of 2, combine the left side with the division law and equate the indices, which gives the quadratic \(x^2 - 2x - 3 = 0\). Both roots are valid, because every value of \(x\) gives positive powers of 2. Common errors are writing \(4^{x^2}\) as \(2^{x^4}\) and forgetting to multiply the whole of \(x + 2\) by 3.

Mark scheme

1 mark
M1 Writes each side as a power of 2 (at least two of \(4^{x^2} = 2^{2x^2}\), \(2^x\) and \(8^{x + 2} = 2^{3x + 6}\) correct) and equates the indices: \(2x^2 - x = 3x + 6\). An answer with no working scores 0 marks.
An answer of \(x = 3\) and \(x = -1\) with no working scores 0 marks: the question says "using algebra". Common errors: \(4^{x^2} = 2^{x^4}\) or \(2^{4x^2}\), and \(8^{x + 2} = 2^{3x + 2}\) (the bracket not multiplied by 3). Taking logarithms of both sides correctly is also M1.
1 mark
A1 \(2x^2 - 4x - 6 = 0\) oe (e.g. \(x^2 - 2x - 3 = 0\))
1 mark
M1 Solves their three-term quadratic by factorising, the formula or completing the square
\((x - 3)(x + 1) = 0\).
1 mark
A1 \(x = 3\) and \(x = -1\)
Both needed, and no others.

Question 4

7 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Given that \(x = 3 - \sqrt{7}\), show that \(x^2 = 6x - 2\).

Hence, or otherwise, find the exact value of
\[x^4 - 6x^3 + 3x^2 + x\]
giving your answer in the form \(a + b\sqrt{7}\), where \(a\) and \(b\) are integers.

Find also the exact value of \(x + \dfrac{2}{x}\).

Model answer

\(x^2 = \left(3 - \sqrt{7}\right)^2 = 9 - 6\sqrt{7} + 7 = 16 - 6\sqrt{7}\)
\(6x - 2 = 6\left(3 - \sqrt{7}\right) - 2 = 18 - 6\sqrt{7} - 2 = 16 - 6\sqrt{7}\)
So \(x^2 = 6x - 2\).
\(x^4 - 6x^3 = x^2\left(x^2 - 6x\right) = x^2 \times (-2) = -2x^2\)
So \(x^4 - 6x^3 + 3x^2 + x = -2x^2 + 3x^2 + x = x^2 + x\)
\(= (6x - 2) + x = 7x - 2\)
\(= 7\left(3 - \sqrt{7}\right) - 2 = 19 - 7\sqrt{7}\)
\(\dfrac{2}{x} = \dfrac{2}{3 - \sqrt{7}} \times \dfrac{3 + \sqrt{7}}{3 + \sqrt{7}} = \dfrac{2\left(3 + \sqrt{7}\right)}{9 - 7} = 3 + \sqrt{7}\)
\(x + \dfrac{2}{x} = 3 - \sqrt{7} + 3 + \sqrt{7} = 6\)

Explanation

Squaring \(3 - \sqrt{7}\) gives \(16 - 6\sqrt{7}\), which equals \(6x - 2\), and this relation reduces the higher powers: \(x^4 - 6x^3 = x^2\left(x^2 - 6x\right) = -2x^2\), so the expression simplifies to \(7x - 2 = 19 - 7\sqrt{7}\). Rationalising \(\dfrac{2}{3 - \sqrt{7}}\) (or dividing \(x^2 - 6x + 2 = 0\) by \(x\)) gives \(x + \dfrac{2}{x} = 6\). Common errors are squaring \(3 - \sqrt{7}\) as \(9 - 7\) and sign slips when the powers are expanded directly.

Mark scheme

1 mark
M1 Squares \(3 - \sqrt{7}\) to three or four terms: \(9 - 6\sqrt{7} + 7\) (or \(9 - 3\sqrt{7} - 3\sqrt{7} + 7\)). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question says that you must show all stages of your working, so \(19 - 7\sqrt{7}\) and 6 written down score nothing. \(\left(3 - \sqrt{7}\right)^2 = 9 - 7\) is M0.
1 mark
A1* \(x^2 = 16 - 6\sqrt{7}\) and \(6x - 2 = 18 - 6\sqrt{7} - 2 = 16 - 6\sqrt{7}\), so \(x^2 = 6x - 2\)
cso. Both sides must be shown to be equal, or \(x^2 - 6x + 2 = 16 - 6\sqrt{7} - 18 + 6\sqrt{7} + 2 = 0\) shown. Solving \(x^2 - 6x + 2 = 0\) to get \(x = 3 \pm \sqrt{7}\) and stating that \(3 - \sqrt{7}\) is a root is also acceptable.
1 mark
M1 Uses \(x^2 = 6x - 2\) to reduce the powers, e.g. \(x^4 - 6x^3 = x^2\left(x^2 - 6x\right) = -2x^2\); or finds \(x^3 = 90 - 34\sqrt{7}\) and \(x^4 = 508 - 192\sqrt{7}\) and substitutes
Alternative: algebraic division of \(x^4 - 6x^3 + 3x^2 + x\) by \(x^2 - 6x + 2\) gives the quotient \(x^2 + 1\) and the remainder \(7x - 2\) (M1 for the division, A1 for the remainder \(7x - 2\)). Direct expansion needs at least one of \(x^3\) and \(x^4\) correct for M1.
1 mark
A1 The expression equals \(x^2 + x\) or \(7x - 2\) (or a correct unsimplified expression in surds, e.g. \(508 - 192\sqrt{7} - 6\left(90 - 34\sqrt{7}\right) + 3\left(16 - 6\sqrt{7}\right) + 3 - \sqrt{7}\))
1 mark
A1 \(19 - 7\sqrt{7}\) (\(a = 19\), \(b = -7\))
cao. Depends on the second M mark. A decimal value (0.479...) scores A0.
1 mark
M1 Rationalises \(\dfrac{2}{3 - \sqrt{7}}\): \(\dfrac{2\left(3 + \sqrt{7}\right)}{9 - 7}\); or divides \(x^2 - 6x + 2 = 0\) by \(x\)
1 mark
A1 \(x + \dfrac{2}{x} = 6\)
cao. From \(\dfrac{2}{x} = 3 + \sqrt{7}\) or from \(x - 6 + \dfrac{2}{x} = 0\).

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