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sums will set you free

how to teach your child numbers arithmetic mathematics

quadratic equations,
model answers



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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 of sums will set you free
how to teach a person number, arithmetic, mathematics on teaching reading

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 quadratic equation. 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) ,
therefore x = 1/3 or - 2 . Ans.
 
(b) Putting the equation into the standard form,
x² - 5x + 2 = 0.
Using the formula quadratic equation
a = 1, b = - 5, c = 2

x = quadratic equation

= equation at abelard.org
Using Square Root Tables

x = equation at abelard.org

= 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
therefore x = quadratic equation

= equation at abelard.org

= equation at abelard.org

= 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

N2 + 2N - 960 = 0 = (N - 30) (N + 32).

As N cannot be negative, N = 30. Ans.

Each passenger paid (£1200/N) + £5 = £45. Ans.

 

 

 

 

 

 

 

 


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Question 4: (January 1971, No. 10, opt.)

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 = equation at abelard.org

Amount of brand B used = equation at abelard.org

therefore equation at abelard.org + 4 = equation at abelard.org,

therefore 1080x - 3 x 1080 + 4x ( x - 3) = 1080x,

therefore 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 = equation at abelard.org

Cost of second journey = equation at abelard.org

therefore 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:
(2x - 5)(2x - 5) = 7 [2]
4x² - 20x + 25 = 7,

subtract 7 from each side:
4x² - 20x + 18 = 0
 
and use ‘the formula’, quadratic equation.

where a = 4, b = 20, c = 18.

x = GCSE quadratic solution

   = GCSE quadratic solution

   = GCSE quadratic solution

   = 3.823 or 1.177 Ans.
  1. 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.
  2. 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.
  3. 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]
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end notes

  1. 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.

  2. 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.
sums will set you free includes the series of documents about economics and money at abelard.org.
moneybookers information e-gold information fiat money and inflation
calculating moving averages the arithmetic of fractional banking

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