A Algebra · sequences and functions — revision question pack

8 specification points · notes, questions, answers and worked methods

A18 · Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Explanation

Rearrange a quadratic equation into $ax^2+bx+c=0$ before choosing a method. Factorising is efficient when integer factors are visible and uses the zero-product rule.
At Higher tier, completing the square rewrites the expression so a square can be isolated, while the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ works generally.
Graphical roots are $x$ -intercepts, or intersection $x$ -coordinates when two graphs are compared.
The examiner expects both solutions unless the context rules one out.
Keep exact surd answers when asked and only round at the final step.

Worked example

Higher tier: Solve $x^2-8x+3=0$ by completing the square.

1.Rewrite the quadratic: $x^2-8x+3=(x-4)^2-13$ .
2.Set it equal to zero: $(x-4)^2=13$ .
3.Take both square roots: $x-4=\pm\sqrt{13}$ .

Answer: $x=4\pm\sqrt{13}$ .

Common mistakes

Don't take only the positive square root and loses one solution.
Don't substitute $b$ into the formula without its sign, so $-b$ is evaluated incorrectly.
Don't round a surd before the final answer and loses accuracy.

Exam tip

Write the substitution into the quadratic formula before evaluating; this is normally where the first method mark is earned.

Tier 1 · Easy

1. Solve $x^2+2x-35=0$ by factorising.[2 marks]

Tier 2 · Standard

1. Higher only: Solve $x^2-8x+3=0$ by completing the square. Give exact answers.[4 marks]

Tier 3 · Hard

1. Higher only: The curves $y=x^2$ and $y=5x+1$ intersect twice. Use the quadratic formula to find the exact $x$ -coordinates, then give the values a graph should show to two decimal places.[5 marks]

A19 · Solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph

Explanation

A simultaneous solution is an ordered pair satisfying both equations; graphically it is an intersection point. For two linear equations, make one variable's coefficients equal or opposite, then add or subtract to eliminate it.
Substitute the first value back to find the second.
At Higher tier, when a line and a quadratic are solved together, substitute the linear expression into the quadratic, solve it, and find the matching second coordinate for every root.
The examiner expects complete ordered pairs and a check in both original equations.
Two $x$ -values alone are incomplete because each belongs to a different point.

Worked example

Solve simultaneously $2x+y=11$ and $x-y=1$ .

1.Add the equations to eliminate $y$ : $3x=12$ .
2.Solve to get $x=4$ .
3.Substitute into $x-y=1$ : $4-y=1$ , so $y=3$ .

Answer: $(x,y)=(4,3)$ .

Common mistakes

Don't add equations whose variable coefficients are not equal or opposite, so no variable is eliminated.
Don't find two possible $x$ -values in a linear-quadratic pair but gives no corresponding $y$ -values.

Exam tip

Write each solution as an ordered pair and substitute both coordinates into both equations when the tariff allows a checking mark.

Tier 1 · Easy

1. Solve simultaneously $y=2x+1$ and $x+y=10$ .[3 marks]

Tier 2 · Standard

1. On the same axes draw $y=0.6x+1$ and $y=5-0.8x$ . Use the graph to estimate their point of intersection to one decimal place.[4 marks]

Tier 3 · Hard

1. Higher only: Solve simultaneously $y=x+1$ and $x^2+y^2=25$ .[5 marks]

A20 · Find approximate solutions to equations numerically using iteration [Higher only]

Explanation

Higher tier only. Iteration rewrites an equation as $x=g(x)$ and repeatedly applies $x_{n+1}=g(x_n)$ from the stated starting value.
Each new output becomes the next input. Keep the calculator's full stored value and round only values you are asked to record, because premature rounding can change later iterates.
Convergence is indicated when successive values agree to the required accuracy, although not every rearrangement converges.
The examiner expects the requested named iterate or a justified approximate root, with enough intermediate values to show the process.
Substitute the final approximation into the original equation as a reasonableness check.

Worked example

The iteration $x_{n+1}=\sqrt{10-x_n}$ starts with $x_0=3$ . Find $x_1$ and $x_2$ to three decimal places.

1. $x_1=\sqrt{10-3}=\sqrt7=2.64575\ldots$ .
2.Use the unrounded value: $x_2=\sqrt{10-2.64575\ldots}=2.71187\ldots$ .
3.Round each requested result to three decimal places.

Answer: $x_1=2.646$ and $x_2=2.712$ .

Common mistakes

Don't substitute $x_0$ again when calculating $x_2$ instead of using $x_1$ .
Don't round each iterate heavily before using it as the next input.

Exam tip

Use the calculator answer key to carry the full previous iterate into the next substitution.

Tier 1 · Easy

1. The iteration $x_{n+1}=\sqrt{10-x_n}$ starts with $x_0=3$ . Work out $x_1$ and $x_2$ , giving each to three decimal places.[2 marks]

Tier 2 · Standard

1. Use $x_{n+1}=\sqrt{(18-x_n)/2}$ with $x_0=3$ to find $x_4$ . Give the result to four decimal places.[4 marks]

Tier 3 · Hard

1. Let $f(x)=x^3+x-12$ . Show that $f(x)=0$ has a root between $2$ and $3$ . Starting with $x_0=2.2$ , use $x_{n+1}=\sqrt[3]{12-x_n}$ to find this root to four decimal places.[6 marks]

A21 · Translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

Explanation

Choose a variable and state exactly what it represents, including units where helpful.
Translate each related quantity into an expression in that variable, then use the stated relationship to form an equation or a pair of simultaneous equations.
Solve using appropriate algebra and interpret the mathematical values in the original situation.
A valid algebraic root may still be impossible as a length, age or count, so reject it with a contextual reason.
The examiner awards marks for forming the model as well as solving it; an unlabelled value of $x$ is not a complete answer when dimensions, prices or numbers of items were requested.

Worked example

A rectangle has width $x$ cm and length $(x+3)$ cm. Its perimeter is $34$ cm. Find both dimensions.

1.Form the perimeter equation: $2x+2(x+3)=34$ .
2.Expand and solve: $4x+6=34$ , so $x=7$ .
3.Interpret the expressions: width $=7$ cm and length $=7+3=10$ cm.

Answer: Width $7$ cm and length $10$ cm.

Common mistakes

Don't form $x+(x+3)=34$ and counts only half of the rectangle's perimeter.
Don't stop at $x=7$ without finding and labelling both requested dimensions.
Don't keep a negative algebraic root even though the quantity is a length or age.

Exam tip

Define the variable before forming the equation and finish with a sentence interpreting every required quantity.

Tier 1 · Easy

1. A rectangle has width $x$ cm and length $(x+3)$ cm. Its perimeter is $34$ cm. Form and solve an equation to find both dimensions.[3 marks]

Tier 2 · Standard

1. A club sells $38$ tickets. Adult tickets cost £7 and junior tickets cost £4. The total received is £203. Form two equations and find how many tickets of each type were sold.[4 marks]

Tier 3 · Hard

1. Mira is $4$ years older than Theo. In $3$ years, the product of their ages will be $192$ . Form an equation and find their current ages.[5 marks]

A22 · Solve linear inequalities in one or two variable(s), and quadratic inequalities in one variable; represent the solution set on a number line, using set notation and on a graph

Explanation

Solve a linear inequality like an equation, reversing the inequality sign only when multiplying or dividing by a negative number.
On a number line, use a filled endpoint when equality is included and an open endpoint for a strict inequality.
At Higher tier, represent two-variable inequalities by drawing each boundary and testing a point to choose the region; strict boundaries are dashed.
For a quadratic inequality, find the roots and test the intervals they define, because the required values may lie inside or outside the roots.
The examiner expects the correct endpoint style, shading and notation as well as the algebraic boundary values.

A strict inequality outside two critical values uses open endpoints and outward shading.

Worked example

Higher tier: Solve $(2x+3)(x-2)>0$ .

1.Find the critical values: $2x+3=0$ gives $x=-\dfrac32$ , and $x-2=0$ gives $x=2$ .
2.Test the three intervals; the product is positive outside the roots.
3.The inequality is strict, so neither endpoint is included.

Answer: $x<-\dfrac32$ or $x>2$ .

Common mistakes

Don't forget to reverse the inequality sign after dividing by a negative number.
Don't use filled endpoints for $<$ or $>$ .
Don't assume a quadratic inequality is always satisfied between its roots without testing signs.

Exam tip

For a graphical answer, the boundary style and the direction of shading are separate marking points.

Tier 1 · Easy

1. Higher only: Solve $3x-7\le11$ . Give the answer in set notation and describe its number-line representation.[3 marks]

Tier 2 · Standard

1. Higher only: On coordinate axes, show the region satisfying both $y\ge2x-1$ and $x+y<5$ . State the intersection of the boundary lines and identify which boundaries are included.[5 marks]

Tier 3 · Hard

1. Higher only: Solve $(2x+3)(x-2)>0$ . Give the solution in set notation and describe it on a number line.[4 marks]

A23 · Generate terms of a sequence from either a term-to-term or a position-to-term rule

Explanation

A term-to-term rule produces each new term from the preceding term or terms, so begin with every stated starting value and apply the operations in the given order.
A position-to-term rule gives a term directly from its position $n$ ; for the first terms substitute $n=1,2,3,\ldots$ unless the question explicitly defines another starting index.
Multi-step, alternating and Fibonacci-type rules require intermediate terms to be shown because a later term may depend on more than one earlier value.
The examiner expects the terms in order and usually gives method credit for correct substitutions or repeated operations even if a later arithmetic error occurs.

Worked example

A sequence starts $2,5$ . Each later term is one more than the sum of the previous two terms. Find the next three terms.

1.Third term: $2+5+1=8$ .
2.Fourth term: $5+8+1=14$ .
3.Fifth term: $8+14+1=23$ .

Answer: $8,14,23$ .

Common mistakes

Don't substitute $n=0$ for the first term when the sequence starts at $n=1$ .
Don't make this mistake: For a two-term recurrence, repeatedly uses the original starting pair instead of the latest two terms.

Exam tip

Write each intermediate term because a correct recurrence method can still earn marks after one arithmetic slip.

Tier 1 · Easy

1. A sequence has position-to-term rule $6n-2$ . Write down its first four terms.[2 marks]

Tier 2 · Standard

1. A sequence starts $2,5$ . Each later term is one more than the sum of the previous two terms. Write down the next three terms.[3 marks]

Tier 3 · Hard

1. The $n$ th term of a sequence is $n^2-3n+4$ . Generate the first six terms.[4 marks]

A24 · Recognise and use triangular, square and cube numbers, arithmetic progressions, Fibonacci type sequences, quadratic sequences, simple geometric progressions (r^n, r rational > 0 or a surd) and others

Explanation

Square and cube numbers have forms $n^2$ and $n^3$ , while triangular numbers have form $\dfrac{n(n+1)}{2}$ . An arithmetic progression has a constant first difference and a quadratic sequence has constant second differences.
Fibonacci-type sequences form later terms from preceding terms.
At Higher tier, a geometric progression has a constant positive rational or surd multiplier and can be written using powers such as $r^n$ .
Check several consecutive steps before identifying a sequence, then use the defining rule consistently.
The examiner expects the named structure to be supported by differences, ratios or a valid term relationship rather than by visual resemblance alone.

Worked example

Higher tier: The sequence $2,2\sqrt3,6,6\sqrt3,\ldots$ is geometric. Find its common ratio and eighth term.

1.Divide consecutive terms: $2\sqrt3\div2=\sqrt3$ , so $r=\sqrt3$ .
2.Use $u_n=2(\sqrt3)^{n-1}$ .
3. $u_8=2(\sqrt3)^7=2(27\sqrt3)=54\sqrt3$ .

Answer: Common ratio $\sqrt3$ ; eighth term $54\sqrt3$ .

Common mistakes

Don't call a sequence arithmetic after checking only one pair of terms.
Don't use first differences to identify a quadratic sequence instead of checking that second differences are constant.
Don't replace the exact surd ratio by a decimal and loses the exact form.

Exam tip

Show a short difference table or two equal consecutive ratios to justify the sequence type.

Tier 1 · Easy

1. The sequence $1,4,9,16,\ldots$ is made from a named type of number. Name the type and write down the next two terms.[2 marks]

Tier 2 · Standard

1. A Fibonacci-type sequence begins $3,7,10,17,27,\ldots$ , with each term after the second equal to the sum of the previous two. Find the eighth term.[3 marks]

Tier 3 · Hard

1. Higher only: A geometric progression begins $2,2\sqrt3,6,6\sqrt3,\ldots$ . State the common ratio and find the eighth term in exact form.[4 marks]

A25 · Deduce expressions to calculate the nth term of linear and quadratic sequences

Explanation

For a linear sequence with common difference $d$ , start with $dn$ and adjust the constant so the expression gives the first term when $n=1$ .
At Higher tier, a quadratic sequence $an^2+bn+c$ has constant second difference $2a$ .
Find $a$ , subtract the values of $an^2$ from the original terms, and identify the linear rule left behind to obtain $b$ and $c$ .
The examiner expects an expression in $n$ , not merely the next term.
Verify the rule against at least two supplied terms; this detects a shifted index or an incorrect constant before the final answer.

Worked example

Higher tier: Find the $n$ th term of $2,7,14,23,34,\ldots$ .

1.First differences are $5,7,9,11$ , so the second difference is $2$ and $a=1$ .
2.Subtract $n^2$ from the terms to get $1,3,5,7,9$ .
3.The remainder has rule $2n-1$ , so combine the parts.

Answer: $n^2+2n-1$ .

Common mistakes

Don't use the constant second difference as $a$ instead of recognising it is $2a$ .
Don't use the first term itself as the constant in a linear $n$ th-term rule.
Don't find a rule that matches one term but does not verify it against later terms.

Exam tip

For a quadratic sequence, display the first and second differences because they secure the method for finding the $n^2$ coefficient.

Tier 1 · Easy

1. Find an expression for the $n$ th term of $7,11,15,19,\ldots$ .[2 marks]

Tier 2 · Standard

1. Higher only: Find an expression for the $n$ th term of $2,7,14,23,34,\ldots$ .[4 marks]

Tier 3 · Hard

1. Higher only: A quadratic sequence starts $4,15,32,55$ . Deduce its $n$ th term and determine the position of the term equal to $207$ .[6 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

A18 · Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

Tier 1 · Easy

1. Answer
- $x=5$ or $x=-7$
Method: $x^2+2x-35=(x+7)(x-5)$ . By the zero-product rule, $x+7=0$ or $x-5=0$ , so $x=-7$ or $x=5$ .

Tier 2 · Standard

1. Answer
- $x=4\pm\sqrt{13}$
Method: $x^2-8x+3=(x-4)^2-13$ . Hence $(x-4)^2=13$ , so $x-4=\pm\sqrt{13}$ and $x=4\pm\sqrt{13}$ .

Tier 3 · Hard

1. Answer
- $x=\frac{5\pm\sqrt{29}}{2}$
- $x\approx-0.19$ or $x\approx5.19$
Method: At an intersection, $x^2=5x+1$ , so $x^2-5x-1=0$ . The formula gives $x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2}=\frac{5\pm\sqrt{29}}{2}$ . These are approximately $-0.19$ and $5.19$ , matching the graph's intersection coordinates.

A19 · Solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph

Tier 1 · Easy

1. Answer
- $x=3$ , $y=7$
Method: Substitute $y=2x+1$ into $x+y=10$ : $x+2x+1=10$ . Hence $3x=9$ , so $x=3$ and $y=2(3)+1=7$ .

Tier 2 · Standard

1. Answer
- $(x,y)\approx(2.9,2.7)$
Method: The solution is where the two lines cross. Reading the graph gives approximately $x=2.9$ and $y=2.7$ ; algebraically the point is $(20/7,19/7)$ , which confirms those graphical estimates.

Tier 3 · Hard

1. Answer
- $(x,y)=(-4,-3)$ or $(3,4)$
Method: Substitute $y=x+1$ into the second equation: $x^2+(x+1)^2=25$ . This simplifies to $2x^2+2x-24=0$ , so $x^2+x-12=0=(x+4)(x-3)$ . Thus $x=-4$ or $x=3$ . Using $y=x+1$ gives $(-4,-3)$ and $(3,4)$ .

A20 · Find approximate solutions to equations numerically using iteration [Higher only]

Tier 1 · Easy

1. Answer
- $x_1=2.646$
- $x_2=2.712$
Method: $x_1=\sqrt{10-3}=\sqrt7=2.64575\ldots$ . Then $x_2=\sqrt{10-2.64575\ldots}=2.71187\ldots$ . To three decimal places these are $2.646$ and $2.712$ .

Tier 2 · Standard

1. Answer
- $x_4=2.7604$
Method: $x_1=2.7386\ldots$ , $x_2=2.7624\ldots$ , $x_3=2.7602\ldots$ and $x_4=2.7604\ldots$ . Retaining the unrounded values at each stage gives $x_4=2.7604$ to four decimal places.

Tier 3 · Hard

1. Answer
- $f(2)=-2$ and $f(3)=18$ , so a root lies between $2$ and $3$
- $x\approx2.1440$
Method: $f(2)=8+2-12=-2$ and $f(3)=27+3-12=18$ , so the sign change shows a root in the interval. The iteration gives $x_1=2.1400\ldots$ , $x_2=2.1443\ldots$ , $x_3=2.1440\ldots$ , $x_4=2.1440\ldots$ and subsequent values remain $2.1440$ to four decimal places.

A21 · Translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

Tier 1 · Easy

1. Answer
- Width $=7$ cm
- Length $=10$ cm
Method: The perimeter equation is $2x+2(x+3)=34$ . Hence $4x+6=34$ , so $4x=28$ and $x=7$ . The width is $7$ cm and the length is $7+3=10$ cm.

Tier 2 · Standard

1. Answer
- $17$ adult tickets and $21$ junior tickets
Method: Let $a$ and $j$ be the numbers of adult and junior tickets. Then $a+j=38$ and $7a+4j=203$ . Subtract $4(a+j)=152$ from the money equation to get $3a=51$ , so $a=17$ and $j=21$ .

Tier 3 · Hard

1. Answer
- Theo is $9$ years old and Mira is $13$ years old.
Method: Let Theo's current age be $x$ , so Mira's is $x+4$ . In three years their ages are $x+3$ and $x+7$ , giving $(x+3)(x+7)=192$ . Thus $x^2+10x-171=0=(x-9)(x+19)$ . The solutions are $x=9$ and $x=-19$ ; reject the negative age. Theo is $9$ and Mira is $13$ .

A22 · Solve linear inequalities in one or two variable(s), and quadratic inequalities in one variable; represent the solution set on a number line, using set notation and on a graph

Tier 1 · Easy

1. Answer
- $\{x:x\le6\}$
- A filled point at $6$ with the line shaded to the left.
Method: Add $7$ to get $3x\le18$ , then divide by $3$ to obtain $x\le6$ . Equality is included, so use a filled point at $6$ and shade all smaller values.

Tier 2 · Standard

1. Answer
- Shade above $y=2x-1$ and below $y=5-x$
- The boundaries meet at $(2,3)$
- $y=2x-1$ is solid and $y=5-x$ is dashed
Method: Draw $y=2x-1$ as a solid line because equality is allowed. Draw $y=5-x$ as a dashed line because $x+y<5$ is strict. Solving $2x-1=5-x$ gives $3x=6$ , so the lines meet at $(2,3)$ . The required region is above the solid line and below the dashed line.

Tier 3 · Hard

1. Answer
- $\{x:x<-\frac32\text{ or }x>2\}$
- Open points at $-\frac32$ and $2$ , shaded outwards.
Method: The critical values are $x=-3/2$ and $x=2$ . The product is positive outside these roots, so $x<-3/2$ or $x>2$ . The inequality is strict, so both endpoints are open and the two outer regions are shaded.

A23 · Generate terms of a sequence from either a term-to-term or a position-to-term rule

Tier 1 · Easy

1. Answer
- $4,10,16,22$
Method: Substitute $n=1,2,3,4$ into $6n-2$ . This gives $4$ , $10$ , $16$ and $22$ .

Tier 2 · Standard

1. Answer
- $8,14,23$
Method: The third term is $2+5+1=8$ . The fourth is $5+8+1=14$ . The fifth is $8+14+1=23$ .

Tier 3 · Hard

1. Answer
- $2,2,4,8,14,22$
Method: Substitute $n=1$ through $6$ : $1-3+4=2$ , $4-6+4=2$ , $9-9+4=4$ , $16-12+4=8$ , $25-15+4=14$ and $36-18+4=22$ .

A24 · Recognise and use triangular, square and cube numbers, arithmetic progressions, Fibonacci type sequences, quadratic sequences, simple geometric progressions (r^n, r rational > 0 or a surd) and others

Tier 1 · Easy

1. Answer
- Square numbers
- $25,36$
Method: The terms are $1^2,2^2,3^2,4^2$ . Therefore they are square numbers and the next terms are $5^2=25$ and $6^2=36$ .

Tier 2 · Standard

1. Answer
- $115$
Method: Continue the rule: the sixth term is $17+27=44$ , the seventh is $27+44=71$ , and the eighth is $44+71=115$ .

Tier 3 · Hard

1. Answer
- Common ratio $=\sqrt3$
- Eighth term $=54\sqrt3$
Method: Each term is multiplied by $\sqrt3$ . The $n$ th term is $2(\sqrt3)^{n-1}$ . For $n=8$ , this is $2(\sqrt3)^7=2(27\sqrt3)=54\sqrt3$ .

A25 · Deduce expressions to calculate the nth term of linear and quadratic sequences

Tier 1 · Easy

1. Answer
- $4n+3$
Method: The common difference is $4$ , so start with $4n$ . This gives $4$ when $n=1$ , which is $3$ below the first term $7$ . Therefore the rule is $4n+3$ .

Tier 2 · Standard

1. Answer
- $n^2+2n-1$
Method: The first differences are $5,7,9,11$ , so the second difference is $2$ and the quadratic begins with $n^2$ . Subtracting $n^2$ gives $1,3,5,7,9$ , whose rule is $2n-1$ . Therefore the $n$ th term is $n^2+2n-1$ .

Tier 3 · Hard

1. Answer
- $3n^2+2n-1$
- $207$ is the eighth term
Method: The first differences are $11,17,23$ , so the second difference is $6$ and $a=3$ . Subtracting $3n^2$ from the terms gives $1,3,5,7$ , whose rule is $2n-1$ . Hence the term is $3n^2+2n-1$ . Set this equal to $207$ : $3n^2+2n-208=0=(3n+26)(n-8)$ . The positive integer solution is $n=8$ .