AQA Level 2 Further Maths · Level 2 Certificate · Chapter 4: Algebra IV

Quadratics, simultaneous equations, the factor theorem, inequalities, indices, proof and sequences

About 90 minutesNo calculator116 marks

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Date

Do Now

The first is what this chapter needs. The rest bring back earlier work so nothing is forgotten.

A1
Needed today
Factorise $6 x^{2} - 5 x - 21$ .
[2 marks]
Answer

A2
From chapter 2
Write $x^{2} + 8 x + 23$ in the form $(x + a)^{2} + b$ .
[2 marks]
Answer
A3
From chapter 3
$f (x) = 3 x - 2$ and $g (x) = x^{2} + 1$ . Work out $g f (- 1)$ .
[2 marks]
Answer
A4
From chapter 1
Expand and simplify $(4 - 7)^{2}$ .
[2 marks]
Answer
A5
GCSE Higher
Solve the simultaneous equations $5 x + 3 y = 9$ and $3 x - 4 y = 17$ .
[2 marks]
Answer $x = \dots\dots$ $y = \dots\dots$

Examples, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it on your own.

Example 1

Solve the simultaneous equations
$x + 2 y - z 2 x - y + 2 z 3 x + y - z = 0 = 13 = 9$

Hint Use two different pairs of equations to remove the same letter,

z

. That leaves two equations in

x

and

y

B1
Solve the simultaneous equations
$2 a + b - c a - 3 b + 2 c 3 a + 2 b + c = 9 = - 11 = 10$
Do not use trial and improvement.
[5 marks]
Answer $a = \dots\dots$ $b = \dots\dots$ $c = \dots\dots$

Example 2

Solve the simultaneous equations $x + y = 3$ and $x^{2} + 2 y^{2} = 18$ .

Hint Make

y

the subject of the linear equation and substitute it into the other one. Find both values of

y

at the end.

B2
Solve the simultaneous equations $y = 2 x + 1$ and $y^{2} - 3 x^{2} = 6$ .
[5 marks]
Answer $(\dots, \dots)$ and $(\dots, \dots)$

Example 3

$f (x) = 2 x^{3} + a x^{2} + b x + 6$ . $(x - 2)$ and $(x + 3)$ are factors of $f (x)$ .

Work out the values of $a$ and $b$ . Hence solve $f (x) = 0$ .

Hint A factor

(x - 2)

means

f (2) = 0

. That gives one equation in

a

and

b

;

(x + 3)

gives another.

B3
$f (x) = 3 x^{3} + a x^{2} + b x - 8$ . $(x + 2)$ and $(x - 1)$ are factors of $f (x)$ .
Work out the values of $a$ and $b$ . Hence solve $f (x) = 0$ .
[5 marks]
Answer $a = \dots$ $b = \dots$ $x = \dots\dots$

Example 4

Solve $2^{2 x + 1} - 9 \times 2^{x} + 4 = 0$ .

Hint

2^{2 x + 1} = 2 \times (2^{x})^{2}

. Let

y = 2^{x}

and solve the quadratic in

y

B4
Solve $3^{2 x + 1} + 8 \times 3^{x} = 3$ .
[4 marks]
Answer $x = \dots\dots$

Practice

Level 2

Each question asks for something different. Show your working.

C1
Solve $3 x^{2} + 6 x - 2 = 0$ .
Give your answers in the form $\frac{a \pm b}{c}$ , where $a$ , $b$ and $c$ are integers.
[3 marks]
Answer

C2
At a cinema, 3 adult tickets and 5 child tickets cost £65.50. 2 adult tickets and 3 child tickets cost £41.50.
Work out the cost of 4 adult tickets and 1 child ticket.
[4 marks]
Answer£ $\dots\dots$
C3
Show that $(2 x - 3)$ is a factor of $2 x^{3} - 3 x^{2} - 8 x + 12$ .
Hence factorise $2 x^{3} - 3 x^{2} - 8 x + 12$ fully.
[4 marks]
Answer
C4
Solve $x (x + 2) ⩾ 3 (2 x + 4)$ .
[3 marks]
Answer $\dots\dots\dots\dots$
C5
$- 2 ⩽ a ⩽ 3$ and $- 5 ⩽ b ⩽ - 1$ .
(a)
Work out the greatest possible value of $ab$ .
[2 marks]
Answer
(b)
Work out the least possible value of $a^{2} - b$ .
[2 marks]
Answer
Total for C5: 4 marks
C6
Write $\frac{( 8 x ^{6} ) ^{\frac{2}{3}}}{2 x ^{- 1}}$ in the form $a x^{n}$ .
Hence solve $\frac{( 8 x ^{6} ) ^{\frac{2}{3}}}{2 x ^{- 1}} = 64$ .
[3 marks]
Answer $\dots\dots\dots$ $x = \dots\dots$
C7
Prove that the sum of the squares of any three consecutive integers is always $2$ more than a multiple of $3$ .
[3 marks]
C8
Work out an expression for the $n$ th term of the quadratic sequence
$2, 3, 10, 23, 42, \dots$
[3 marks]
Answer
C9
The curve $y = a x^{2} + b x + c$ passes through the points $(- 1, 1)$ , $(2, 4)$ and $(3, 17)$ .
Work out the values of $a$ , $b$ and $c$ .
[4 marks]
Answer $a = \dots\dots$ $b = \dots\dots$ $c = \dots\dots$

Exam-style questions

AQA exam style

Answer every question in the space given. Show your working: most of the marks are for method.

D1
Solve $\frac{6}{x + 2} + \frac{1}{x - 3} = 2$
You must show your working.
[4 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$

D2
$f (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ . Kai works out $f (2) = 0$ and says, "So $x = - 2$ is a solution of $f (x) = 0$ ."
Kai is wrong.
(a)
Show that $x = - 2$ is not a solution of $f (x) = 0$ .
[1 mark]
(b)
Write down the factor of $f (x)$ that $f (2) = 0$ shows.
[1 mark]
Answer
(c)
Hence solve $f (x) = 0$ . You must show your working.
[4 marks]
Answer $x = \dots\dots\dots\dots\dots\dots$
Total for D2: 6 marks
D3
Work out the integers $x$ that satisfy both
$4 x - 1 > x + 2 and 2 x^{2} - 3 x - 9 ⩽ 0$
You must show your working.
[4 marks]
Answer $\dots\dots\dots\dots$
D4
Here is an expression.
$\frac{9 ^{x}}{2 7 ^{x - 1}}$
(a)
Write the expression in the form $3^{k}$ , where $k$ is an expression in $x$ .
[2 marks]
Answer
(b)
Hence solve $\frac{9 ^{x}}{2 7 ^{x - 1}} = \frac{1}{9}$
[2 marks]
Answer $x = \dots\dots$
Total for D4: 4 marks
D5
Prove that the equation $x^{2} + k x + k - 2 = 0$ has two different real roots for every value of $k$ .
[4 marks]
D6
The $n$ th term of a sequence is $\frac{6 n - 1}{2 n + 3}$
(a)
A term of the sequence has the value $2.6$ . Work out which term it is.
[2 marks]
Answer
(b)
Write down the limiting value of the sequence as $n \to \infty$ .
[1 mark]
Answer
(c)
Show that every term of the sequence is less than $3$ .
[2 marks]
Total for D6: 5 marks
D7
The $n$ th term of a quadratic sequence is $a n^{2} + bn + c$ .
The 1st term is $5$ , the 3rd term is $25$ and the 5th term is $69$ .
(a)
Work out an expression for the $n$ th term. Do not use trial and improvement.
[4 marks]
Answer
(b)
Show that $265$ is not a term of the sequence.
[3 marks]
Total for D7: 7 marks
D8
A rectangle has a perimeter of $82$ cm. Each diagonal of the rectangle is $29$ cm long.
Work out the area of the rectangle. You must show your working.
[5 marks]
Not drawn accurately
Answer $\dots\dots$ cm²
D9
Circle the value of $6 4^{- \frac{2}{3}}$
$- 16 \frac{1}{16} - \frac{128}{3} \frac{1}{4}$
[1 mark]

Extension

Stretch

No route is given. Spend five minutes on a problem before you look at a hint. The hints are on the last page of the worksheet, and you can use one at a time.

E1
Solve $2 x^{\frac{1}{2}} - 7 x^{\frac{1}{4}} + 3 = 0$ .
[4 marks]
Hint 1 · What to try
$x^{\frac{1}{2}} = (x^{\frac{1}{4}})^{2}$ . Give $x^{\frac{1}{4}}$ a letter of its own.
Hint 2 · The first line
With $y = x^{\frac{1}{4}}$ : $2 y^{2} - 7 y + 3 = 0$ , so $(2 y - 1) (y - 3) = 0$ .
Hint 3 · The full method
$x^{\frac{1}{4}} = \frac{1}{2}$ gives $x = (\frac{1}{2})^{4} = \frac{1}{16}$ , and $x^{\frac{1}{4}} = 3$ gives $x = 3^{4} = 81$ .
Answer $x = \dots\dots$ or $x = \dots\dots$

E2
The line $y = k x + 3$ is a tangent to the curve $y = x^{2} - 3 x + 12$ .
Work out the two possible values of $k$ .
[4 marks]
Hint 1 · What to try
Put the two expressions for $y$ equal. A tangent meets the curve at exactly one point.
Hint 2 · The first line
$x^{2} - (3 + k) x + 9 = 0$ must have one repeated root.
Hint 3 · The full method
The discriminant is $0$ : $(3 + k)^{2} - 36 = 0$ , so $3 + k = 6$ or $3 + k = - 6$ , which gives $k = 3$ or $k = - 9$ .
Answer $k = \dots\dots$ or $k = \dots\dots$
E3
Prove that $n^{3} - n$ is a multiple of $6$ for every positive integer $n$ .
[3 marks]
Hint 1 · What to try
Factorise $n^{3} - n$ fully.
Hint 2 · The first line
$n^{3} - n = (n - 1) n (n + 1)$ , three consecutive integers.
Hint 3 · The full method
Of any two consecutive integers one is even, so the product is a multiple of $2$ . Of any three consecutive integers one is a multiple of $3$ . So the product is a multiple of $2 \times 3 = 6$ .
E4
$f (x) = x^{3} + a x^{2} + b x + c$ . $(x - 1)$ and $(x + 2)$ are factors of $f (x)$ , and $f (2) = 20$ .
Work out the values of $a$ , $b$ and $c$ , and factorise $f (x)$ fully.
[5 marks]
Hint 1 · What to try
Each fact gives one equation in $a$ , $b$ and $c$ .
Hint 2 · The first line
$1 + a + b + c = 0$ , $- 8 + 4 a - 2 b + c = 0$ , $8 + 4 a + 2 b + c = 20$ .
Hint 3 · The full method
The last two give $4 b = 4$ , so $b = 1$ . Then $a + c = - 2$ and $4 a + c = 10$ , so $a = 4$ and $c = - 6$ . $f (x) = x^{3} + 4 x^{2} + x - 6 = (x - 1) (x + 2) (x + 3)$ .
Answer $a = \dots$ $b = \dots$ $c = \dots$ $f (x) = \dots\dots\dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve a quadratic equation, including one that starts as algebraic fractions.
I can solve a linear and a quadratic equation together.
I can use the factor theorem to find factors, unknown coefficients and roots.
I can solve linear and quadratic inequalities and combine ranges.
I can work with fractional and negative indices and solve equations in powers.
I can write an algebraic proof.
I can find the $n$ th term and the limiting value of a sequence.
I can solve three equations in three unknowns.

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