OxfordAQA International AS and A-level Mathematics: Surds, indices, quadratics, simultaneous equations (9660) – Revision Notes

Condensed recall for specification content P1.1 (Algebra) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification states this content is part of International AS Unit P1 (Pure Maths, section 3.1). The study guide, with full explanations and worked examples, sits behind these notes. Then test yourself with the practice questions.

Course links: OxfordAQA Mathematics hub, printable checklist, free 10-minute diagnostics and the Pure Maths 1 revision notes for the rest of the unit.

Key results at a glance

Topic	Result to remember
Surd products	√(ab) = √a × √b; √a × √a = a
Simplifying	Pull out the largest square factor: √48 = 4√3
Rationalising √a	Multiply top and bottom by √a
Rationalising p + √q	Multiply by p − √q; the denominator becomes p² − q
Index laws	add exponents to multiply, subtract to divide, multiply for a power of a power
Special powers	a⁰ = 1; a⁻ⁿ = 1/aⁿ; a^(1/n) is the nth root of a; a^(m/n) is that root raised to m
Completing the square	x² + bx + c = (x + b/2)² + (c − b²/4)
Vertex of a(x + p)² + q	(−p, q); line of symmetry x = −p
Formula gives	x = [−b ± √(b² − 4ac)] / (2a)
Discriminant	b² − 4ac

Surds: method in steps

1. Simplify each surd by pulling out the largest square factor.
2. Collect like surds (same number under the root).
3. When expanding, remember √a × √a = a, so the surd term can vanish.
4. To rationalise, choose the conjugate (change the sign between the terms).

Reminder example. Rationalise 6/(2 + √3).

6/(2 + √3) × (2 − √3)/(2 − √3) = 6(2 − √3)/(4 − 3) = 12 − 6√3

Check: 12 − 6√3 ≈ 1.61, and 6/(2 + 1.732) ≈ 1.61.

Reminder example. Express (√3 − 1)² in the form a + b√3.

3 − 2√3 + 1 = 4 − 2√3

Indices: method in steps

1. Rewrite roots as powers, and put numbers over a common base if you are solving.
2. Deal with the negative sign first (reciprocal), then the root, then the power.
3. For a fraction in brackets, invert it when the power is negative.

Expression	Working	Value
16^(3/4)	(⁴√16)³ = 2³	8
(1/4)^(−3/2)	4^(3/2) = (√4)³	8
125^(−2/3)	1/(∛125)² = 1/25	1/25

To solve 2^(x + 3) = 32, write 32 = 2⁵, so x + 3 = 5 and x = 2. This works because equal bases need equal exponents.

Quadratics: which method?

Situation	Best choice
Integer roots, small coefficients	Factorise
a ≠ 1, factors not obvious	Split the middle term with a pair of numbers whose product is ac and sum is b
Exact answer wanted, or vertex needed	Complete the square
Messy coefficients	Formula
Question says “leave in surd form”	Formula or completing the square

Completing the square in steps (for ax² + bx + c):

1. Take out a from the x² and x terms: a(x² + (b/a)x) + c.
2. Inside the bracket, halve the x coefficient and subtract the square of that half.
3. Write a[(x + b/2a)² − (b/2a)²] + c and tidy up.

Reminder example. x² + 14x + 10 = (x + 7)² − 39. The vertex is (−7, −39), the line of symmetry is x = −7, and the minimum value of y is −39.

If a < 0 the parabola opens downward and the vertex is a maximum.

Discriminant: must-know table

b² − 4ac	Number of real roots	Graph and the x-axis
> 0	two distinct	meets the axis at two points
= 0	one repeated (equal)	vertex lies on the axis
< 0	none	curve misses the axis

Reminder example. For x² − 4x + k = 0, b² − 4ac = 16 − 4k. Equal roots: k = 4. Distinct real roots: k < 4. No real roots: k > 4.

When a question says “show that the equation has real roots for all k”, write the discriminant as a perfect square or a sum of a square and a positive number.

Simultaneous equations (linear and quadratic)

1. Rearrange the linear equation to isolate x or y.
2. Put that into the quadratic and collect terms into a single quadratic.
3. Solve it (usually two values).
4. Return to the linear equation to get the partner of each value.
5. Write the answers as pairs.

Reminder example. y = x + 2 and y = x². Then x² = x + 2, so x² − x − 2 = 0, giving (x − 2)(x + 1) = 0. The solutions are (2, 4) and (−1, 1).

Geometrically, the solutions are where the line meets the curve. One solution means a tangent; none means no intersection.

Inequalities: must-know distinctions

Type	Rule
Linear	Rearrange as usual, but turn the sign round when multiplying or dividing by a negative
Quadratic, a > 0, “> 0”	Outside the roots: x < α or x > β
Quadratic, a > 0, “< 0”	Between the roots: α < x < β
x² < k (k > 0)	−√k < x < √k
x² > k (k > 0)	x < −√k or x > √k

For a quadratic inequality, always rearrange to have 0 on one side, find the critical values, then picture or sketch the parabola. Never divide both sides by an expression that contains x without considering its sign.

Reading a parabola from its equation

Feature	How to find it
Direction	Sign of a: upward if a > 0, downward if a < 0
y-intercept	Put x = 0, giving the constant term c
x-intercepts	Solve the equation; the discriminant tells you how many
Line of symmetry	x = −b/2a, or x = −p from the completed square
Vertex	Substitute the line of symmetry into the equation, or read (−p, q)

Example: y = −x² + 6x − 5 = −(x − 3)² + 4. The parabola opens downward, the vertex (3, 4) is a maximum, the line of symmetry is x = 3, the y-intercept is −5, and the roots are x = 1 and x = 5.

Index traps

• (2x)³ = 8x³, not 2x³. The power applies to the number and the letter.
• 3x⁻² means 3 × (1/x²), so it equals 3/x², not 1/(3x²).
• √x = x^(1/2) and 1/√x = x^(−1/2). Convert before you multiply or divide.
• a^(m/n) can be worked as the nth root of a, raised to m, or the nth root of aᵐ. The first keeps numbers small.
• 2⁻³ = 1/8. A negative power never makes the value negative.

Two combined reminders

Solve by completing the square, exact answer. Solve 2x² − 4x − 3 = 0.

2(x² − 2x) − 3 = 0
2[(x − 1)² − 1] − 3 = 0
2(x − 1)² − 5 = 0
(x − 1)² = 5/2
x = 1 ± √(5/2) = 1 ± √10/2

The step √(5/2) = √10/2 is a rationalisation. A tidy final surd is easier to check against a decimal.

Tangent condition. Find k so that the line y = x + k touches the curve y = x² − 5x + 11.

x + k = x² − 5x + 11
x² − 6x + (11 − k) = 0
b² − 4ac = 36 − 4(11 − k) = 4k − 8
Touching means exactly one solution: 4k − 8 = 0, so k = 2
With k = 2 the equation is (x − 3)² = 0, so x = 3 and y = 5

The line meets the curve at one point only, (3, 5). A value of k above 2 gives two intersections; below 2 gives none.

Speed checks for the exam

• After rationalising, estimate both forms as decimals. They must agree.
• After factorising, expand mentally to confirm the middle term.
• After solving a quadratic inequality, test one value from each region in the original inequality.
• After solving simultaneous equations, put both pairs into the equation you did not use for substitution.
• For the discriminant, write a, b and c in a list before you substitute, especially when b or c is negative.

Quick self-test

1. Simplify √75 − √12.
2. Rationalise 1/(√5 − √3).
3. Simplify (√3 − 1)² in the form a + b√3.
4. Evaluate 16^(3/4).
5. Evaluate (1/4)^(−3/2).
6. Solve 2^(x + 3) = 32.
7. Express x² + 14x + 10 as (x + p)² + q and give the vertex.
8. Find the values of k for which x² − 4x + k = 0 has no real roots.
9. Solve x² − 5x + 6 ≤ 0.
10. Solve y = x + 2 and y = x² simultaneously.
11. Solve x² + x − 12 > 0.
12. Solve −2x > 6.

Answers

1. 5√3 − 2√3 = 3√3
2. (√5 + √3)/(5 − 3) = (√5 + √3)/2
3. 4 − 2√3
4. 8
5. 8
6. x = 2
7. (x + 7)² − 39, vertex (−7, −39)
8. 16 − 4k < 0, so k > 4
9. (x − 2)(x − 3) ≤ 0, so 2 ≤ x ≤ 3
10. (2, 4) and (−1, 1)
11. The product (x + 4)(x − 3) is positive outside its roots: x < −4 or x > 3
12. x < −3

Where marks are usually lost

• Leaving a surd in the denominator when the question says “rationalise” or “exact form”.
• Errors in the sign of the conjugate or in squaring (√6 + √2)² as 6 + 2 only.
• Applying the power before the root when numbers get large, then making an arithmetic slip.
• Treating a⁻ⁿ as negative.
• Completing the square without taking out the factor a first.
• Stating the vertex as (p, q) instead of (−p, q) from a(x + p)² + q.
• Writing a discriminant condition as an equation when the question asks for a range.
• Solving a quadratic inequality as if it were an equation and giving only the critical values.
• Giving only the x-values for simultaneous equations when the question asks for coordinates or pairs.
• Rounding too early in a multi-step question that asks for an exact value.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Content P1.1 (Algebra), International AS Unit P1 (Pure Maths).