how to teach your child number
arithmetic mathematics - quadratic equations, model
answers is part of the
series of documents about fundamental education at abelard.org. These
pages are a sub-set ofsums will
set you free
how to teach a person number,
arithmetic, mathematics
This page is not part of the set of
pages on how to teach basic sums/maths/reasoning. This
page is to compare changes in GCE/GCSE exam papers between 1970 and 2010.
(GCE/GCSE exams are taken in the UK, normally taken at
age 16.) Of course, this page can be used for the study
of exam questions of this type.
GCE Mathematics
question, 1970
The following model answers are taken from G.C.E.
Model Answers for ‘O’ Level Mathematics,
published by Artemis Press, 9th edition 1973. The questions
taken from papers for Syllabus D of the University of
London G.C.E. examination.
From the General Notes:
The examination consists of three papers. Paper 1
is multiple choice, while Papers 2 & 3 are of two
hours. These latter have 8 obligatory questions, and
a choice of four out of six optional questions. It is
recommended that 6 minutes are spent on each obligatiory
question and 15 minutes on each optional one. This allows
‘some’ time for revision.
There are eight obligatory questions (section A),
each carrying 3 to 6 marks, making a total of 32 marks
for section A of each paper.
Optional questions (section B) carry 17 marks each.
Candidates answer four questions from six. Total marks
for section B is 68.
Iincidentally, log tables are provided and slide
rules may be used.
Work must be followable (layout, clarity of reasoning
etc.)
Right answers alone count for very little, the answer
must be set out clearly, including with correct punctuation
and grammar!
The answer must be able to make sense if read out
loud.
Section XI : Solution of Quadratic Equations
[By suitable rearrangement, it is always possible
to put any second-degree or quadratic equation into the
"standard" form ax² + bx + c =
0. The values x which will satisfy this equation are called
its roots. If the left-hand side can be factorised, each
factor separately equated to 0 will give one of the two
roots. If both factors are the same, the roots are the
same. If factorisation is difficult or impossible, the
solution may be obtained by substitution in the formula .
This formula is the result of applying the method of "completing
the square" to form ax² + bx + c =
0.]
Question 1: (January 1969,
No. 3-ii, oblig.)
Solve the equations
(a) 3x² + 5x = 2,
(b) x² + 2 = 5x.
Give the solutions correct to two places of decimals where
necessary.
Answer:
(a) 3x² + 5x - 2 = 0
= (3x - 1) (x + 2) , x = 1/3 or - 2 . Ans.
(b) Putting the equation into the standard form,
x² - 5x + 2 = 0.
Using the formula
a = 1, b = - 5, c = 2
x =
=
Using Square Root Tables
x =
= 4·56 or 0·44. Ans.
[Note: The above answer is written out a little more
fully than is necessary; cf. the next example.]
Question 2: (Summer 1971,
Paper II, No. 2, oblig.)
Solve the equation x2
+ 3x - 1 = 0, giving the roots correct to two
decimal places.
Answer : x² + 3x - 1 = 0.
a = 1, b = 3, c = - 1 x =
=
=
= 0·30 or - 3·30. Ans
Question 3: (Summer 1971,
Paper III, No. 10, opt.)
A travel agency chartered a plane for £1200 to
fly a party of people abroad, it being agreed that each
member of the party should pay an equal share of the cost.
It was later discovered that 4 members of the party
would be unable to travel. The agency calculated that
if it contributed £30, the fare of each of the remaining
passengers would have to be increased by £5 in order
to Cover the cost of £1200.
If N represents the original number of people in the
party
(a) write down an expression
for the number of pounds each passenger would have paid
originally,
(b) form an equation in N
and solve it.
Hence calculate the amount each passenger finally paid.
Answer: (a) Each of the original passengers
would have paid
£1200/N. Ans
(b) Equating the loss of four
fares to the sum needed to make good the loss,
4 x £1200/N = £5 (N - 4) + £30.
Multiplying through by N, dividing by 5 and rearranging
gives
A motorist makes a journey of 1080 miles using petrol
brand A at an average rate of x miles per gallon. He changes
to a cheaper petrol, brand B, with which his car travels
3 miles less per gallon, and uses 4 gallons more for the
same journey. Form an equation for x and solve it to find
the number of miles per gallon with brand A.
Given that brand A costs 35p per gallon and that brand
B costs 32tp per gallon, calculate the difference in cost
of the two journeys.
Answer:
Amount of brand A used =
Amount of brand B used =
+ 4 =
,
1080x - 3 x 1080 + 4x ( x -
3) = 1080x,
x² - 3x -3 x 270 = 0
= (x + 27) (x - 30).
As x must be positive, x = 30 miles per gallon. Ans.
Cost of first journey =
Cost of second journey =
difference in cost = 40p. Ans.
GCSE Mathematics
question, 2010
Solve the equation
(2x-5)² = 7
Answer:
Square-root both sides:
2x - 5 = √7
Add 5 to each side:
2x = √7 + 5
Divide by 2:
x = (√7 + 5) / 2
= 7.646 / 2, or 2.354 / 2
= 3.823 or 1.177 Ans. [1]
Alternative solution:
Multiply out the left-hand side:
It is always easier to solve an equation given in
explicit form than to solve a “word problem”
which must first be interpreted into the form of the
same equation.
When a model answer starts part-way through a potential
solution, able students will recognise this and proceed
from there, so for them the question is easier. Less
able students will instead convert back to the standard
form and solve that, so for them the question is harder.
There is likely also to be a group of students who
will recognise the easier solution but worry about whether
it is acceptable, and thus go the harder route anyway.
[Andy
Walker]
end notes
When a square root is resolved
[worked out], there are always two answers, one positive
and one negative. For instance, (- 2) x (- 2) = 4, (+2)
x (+ 2) = 2 x 2 = 4. (See also multiplying
positive numbers and negative numbers.) Thus the
square root of any number x² is either
or both +x and/or -x. When solving
a real world problem that results in a quadratic equation,
it is often easy to see whether one or other or both
of the solutions are relevant in the real world. See,
for example, question
4 above.
When a square such as
(2x-5)² is multipled out, that is the
sum ( 2x-5)( 2x-5), the result takes
the form (a² + 2ab + b²). For ( 2x-5)²,
a = 2x and b = 5; so (2x-5)²
= 4x² + 20x + 25.
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set you free includes the series
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