Algebraic fractions, completing the square & algebraic proof – IGCSE Maths exam questions with mark scheme

1Grade 6–7HigherCalculator4 marks

Leo is asked to simplify $\dfrac{x^2+3x}{x^2+9}$
He writes $\dfrac{x^2+3x}{x^2+9} = \dfrac{3x}{9} = \dfrac{x}{3}$

(a) Explain what Leo has done wrong. [1]

(b) Simplify fully $\dfrac{x^2-9}{2x^2+5x-3}$ [3]

(a) explanation that $x^2$ is a term, not a factor, so it cannot be cancelled, e.g. 'you can only cancel common factors, not terms that are added' (accept 'he cancelled $x^2$ from a sum'; ignore any attempt to re-simplify)B1
(b) $x^2-9 = (x-3)(x+3)$M1
(b) $2x^2+5x-3=(2x-1)(x+3)$ (or a pair of brackets that expands to give 2 of the 3 terms correct)M1
(b) $\dfrac{x-3}{2x-1}$ caoA1
Answer: (a) He cancelled terms ($x^2$) instead of common factors. (b) $\dfrac{x-3}{2x-1}$
Examiner tip: Factorise top and bottom completely before cancelling: only whole brackets (factors) can be cancelled.

2Grade 6–7HigherCalculator3 marks

(a) Write $x^2+10x+7$ in the form $(x+a)^2+b$, where $a$ and $b$ are integers. [2]

(b) Hence write down the coordinates of the turning point of the curve with equation $y = x^2+10x+7$ [1]

(a) $(x+5)^2$ seen, or $a=5$M1
(a) $(x+5)^2-18$ cao (accept $a=5,\ b=-18$)A1
(b) $(-5,\,-18)$ (ft their $(-a,\ b)$ from (a))B1ft
Answer: (a) $(x+5)^2-18$ (b) $(-5, -18)$
Examiner tip: Halve the $x$-coefficient for $a$, then subtract $a^2$: $7-25=-18$, not $7+25$.

3Grade 6–7HigherCalculator3 marks

Write $\dfrac{3}{x+2} - \dfrac{2}{x-1}$ as a single fraction in its simplest form. [3]

for a common denominator of $(x+2)(x-1)$ with at least one numerator correct, e.g. $\dfrac{3(x-1)}{(x+2)(x-1)}$M1
for $\dfrac{3(x-1)-2(x+2)}{(x+2)(x-1)}$ oe (brackets must be seen or implied by $3x-3-2x-4$)M1
$\dfrac{x-7}{(x+2)(x-1)}$ (accept denominator $x^2+x-2$)A1
Answer: $\dfrac{x-7}{(x+2)(x-1)}$
Examiner tip: The minus sign applies to the whole of $2(x+2)$: $-2x-4$, not $-2x+4$.

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4Grade 8–9HigherCalculator4 marks

Solve $\dfrac{4}{x+1} + \dfrac{3}{2x-1} = 2$
Show clear algebraic working and give your solutions correct to 3 significant figures. [4]

for multiplying through by $(x+1)(2x-1)$, e.g. $4(2x-1)+3(x+1)=2(x+1)(2x-1)$M1
for reducing to a 3-term quadratic, e.g. $4x^2-9x-1=0$ oeM1
for correct substitution into the quadratic formula for their 3-term quadratic, e.g. $\dfrac{9\pm\sqrt{(-9)^2-4\times4\times(-1)}}{2\times4}$ (or correct completing the square)M1
$x=2.36$ and $x=-0.106$ (both required; accept $\dfrac{9\pm\sqrt{97}}{8}$ followed by values to at least 3 s.f.)A1
Answer: $x = 2.36$ or $x=-0.106$ (3 s.f.)
Examiner tip: Multiply every term, including the 2 on the right, by both denominators.

5Grade 6–7HigherCalculator3 marks

$n$ is a positive integer.
Prove, using algebra, that $(2n+3)^2-(2n-3)^2$ is always a multiple of 12 [3]

for expanding both brackets with at least 3 of the 6 terms correct, e.g. $4n^2+12n+9-(4n^2-12n+9)$ (or using difference of two squares: $(2n+3+2n-3)(2n+3-2n+3)$)M1
for $24n$ (or $4n\times6$)A1
for a complete argument, e.g. $24n = 12\times2n$ and $2n$ is an integer, so it is a multiple of 12C1
Answer: $(2n+3)^2-(2n-3)^2 = 24n = 12(2n)$, a multiple of 12
Examiner tip: Testing numbers is not a proof: you must reach $12\times(\text{an integer})$ in algebra and say so.

6Grade 8–9HigherCalculator4 marks

(a) Write $2x^2-12x+23$ in the form $a(x+b)^2+c$, where $a$, $b$ and $c$ are integers. [3]

(b) Using your answer to part (a), explain why the equation $2x^2-12x+23=0$ has no real solutions. [1]

(a) for $2(x^2-6x)$ or $2\left(x^2-6x+\tfrac{23}{2}\right)$ oeM1
(a) for $2(x-3)^2$ seen, or $a=2,\ b=-3$M1
(a) $2(x-3)^2+5$ caoA1
(b) explanation, e.g. $2(x-3)^2\ge0$ so $2(x-3)^2+5\ge5$, so it can never equal 0 (ft their $c>0$; accept 'minimum value is 5')C1ft
Answer: (a) $2(x-3)^2+5$ (b) $2(x-3)^2 \ge 0$, so the expression is always at least 5 and never 0
Examiner tip: When you take out the factor 2, the correction is $2\times9=18$, not 9: $23-18=5$.

7Grade 8–9HigherCalculator4 marks

Prove that the sum of the squares of any three consecutive integers is never a multiple of 3 [4]

for representing three consecutive integers algebraically, e.g. $n-1,\ n,\ n+1$ or $n,\ n+1,\ n+2$M1
for squaring and adding with at least 2 of the expansions correct, e.g. $n^2-2n+1+n^2+n^2+2n+1$M1
for $3n^2+2$ (or $3n^2+6n+5$ for $n,\,n+1,\,n+2$)A1
for a complete conclusion, e.g. $3n^2$ is a multiple of 3 so $3n^2+2$ is 2 more than a multiple of 3 (or $3(n^2+2n+1)+2$), so it is never a multiple of 3C1
Answer: $(n-1)^2+n^2+(n+1)^2 = 3n^2+2$, which leaves remainder 2 when divided by 3
Examiner tip: Getting $3n^2+2$ is not enough for full marks: state that 2 more than a multiple of 3 cannot be a multiple of 3.

8Grade 8–9HigherCalculator4 marks

Simplify fully $\dfrac{x^2-4x}{x^2-x-12} \div \dfrac{2x^2}{x^2+2x-3}$ [4]

for inverting the second fraction and multiplying, e.g. $\dfrac{x^2-4x}{x^2-x-12}\times\dfrac{x^2+2x-3}{2x^2}$M1
for factorising at least two of $x^2-4x=x(x-4)$, $x^2-x-12=(x-4)(x+3)$, $x^2+2x-3=(x+3)(x-1)$M1
for all three factorised correctly with cancelling of a common factor shownM1
$\dfrac{x-1}{2x}$ cao (accept $\dfrac{1}{2}-\dfrac{1}{2x}$)A1
Answer: $\dfrac{x-1}{2x}$
Examiner tip: Flip the second fraction first, then factorise everything, then cancel; don't leave an $x$ on top and bottom.

9Grade 6–7HigherCalculator4 marks

The diagram shows a sketch of the curve with equation $y = x^2+px+q$, where $p$ and $q$ are integers.
The turning point of the curve is $(3,\,-4)$.

(a) Write the equation of the curve in the form $y=(x+a)^2+b$ [2]

(b) Find the value of $p$ and the value of $q$. [2]

(a) $(x-3)^2$ seen, or $a=-3$B1
(a) $y=(x-3)^2-4$B1
(b) for expanding their $(x-3)^2-4$ with at least 2 terms correct, e.g. $x^2-6x+9-4$M1
(b) $p=-6,\ q=5$ (both; ft their (a))A1ft
Answer: (a) $y=(x-3)^2-4$ (b) $p=-6$, $q=5$
Examiner tip: A turning point at $x=3$ means the bracket is $(x-3)$, not $(x+3)$.

10Grade 6–7HigherCalculator3 marks

Prove that the product of any two consecutive odd numbers is always 1 less than a multiple of 4 [3]

for two consecutive odd numbers in algebra and a product, e.g. $(2n+1)(2n+3)$ (accept $(2n-1)(2n+1)$)M1
for $4n^2+8n+3$ (or $4n^2-1$)A1
for a complete argument, e.g. $4n^2+8n+3 = 4(n^2+2n+1)-1$, which is 1 less than a multiple of 4 (or $4n^2-1$ is one less than $4\times n^2$)C1
Answer: e.g. $(2n+1)(2n+3)=4n^2+8n+3=4(n^2+2n+1)-1$
Examiner tip: Consecutive odd numbers differ by 2: use $2n+1$ and $2n+3$, not $2n+1$ and $2n+2$.

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