CAIE A-Level Mathematics 1.1 Quadratics Question Bank

Practise completing the square, solving quadratic equations and inequalities, using the discriminant, eliminating simultaneous variables and recognising quadratic form.

Exam points

complete the square to expose the vertex, range or exact roots of a quadratic
use b² − 4ac with strict or non-strict inequalities to classify real roots
substitute or eliminate variables and solve the resulting quadratic for every valid solution

Covered syllabus points

1.1.1—Completing the square 1.1.2—Quadratic discriminant 1.1.3—Quadratic equations 1.1.4—Simultaneous equations 1.1.5—Quadratic-form equations

[Maximum number: 5]

(a)

Express $3 y^{2}-12 y-15$ in the form $3(y+a)^{2}+b$ , where a and b are constants.

[ 2 ]

$3(y-2)^{2}-27$ or a=-2, b=-27

B1 B1

(b)

Hence find the exact solutions of the equation $3 x^{4}-12 x^{2}-15=0$ .

[ 3 ]

$\left(x^{2}-2\right)^{2}=9$ leading to $x^{2}-2= \pm 3$

M1

Must be $x^{2}$ unless substitution is clear.

$x^{2}=-1$ or $x^{2}=5$

M1

Marking guidance:

Allow omission of -1 if $\pm 3$ seen.

$x= \pm \sqrt{5}$

A1

B1 SC if M1M1 not awarded. Ignore $\pm \mathrm{i}, \mathrm{i},-\mathrm{i}, \sqrt{ }-1$ .
Use of calculator with no working scores 0/3.

Alternative method for Question 1(b)

$3 x^{4}-12 x^{2}-15=0$ leading to $3\left(x^{2}-5\right)\left(x^{2}+1\right)[=0]$

(M1)

$x^{2}=-1$ or $x^{2}=5$

(M1)

Allow omission of -1 if factors seen. Factorising or other
valid method.

$x= \pm \sqrt{5}$

(A1)

B1 SC if M1M1 not scored. Ignore $\pm \mathrm{i}, \mathrm{i},-\mathrm{i}, \sqrt{ }-1$ .
Use of calculator with no working scores 0/3.

[Maximum number: 6]

The function f is defined, for $x \in \mathbb{R}$ , by $\mathrm{f}: x \mapsto x^{2}+a x+b$ , where a and b are constants.

(a)

It is given instead that a=5 and that the roots of the equation f(x)=0 are k and -2 k, where k is a constant.

Find the values of b and k.

[ 3 ]

$(x-k)(x+2k)\equiv x^2+5x+b.$
M1

Comparing coefficients of x:
k=5.
A1

$b=-2k^2=-50.$
A1

(b)

Show that if the equation f(x+a)=a has no real roots then $a^{2}<4(b-a)$ .

[ 3 ]

$(x+a)^2+a(x+a)+b=a.$
M1*

Use the discriminant for the resulting quadratic:
$9a^2-4(2a^2+b-a)<0.$
DM1

Hence
$a^2<4(b-a).$
A1 AG

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Covered syllabus points

1.1 Quadratics question 1

Question (a)

Question (b)

1.1 Quadratics question 2

Question (a)

Question (b)