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April 2001

Number 15

Maths for Physics: Algebraic Manipulation II This Factsheet will explain how to solve quadratic equations and simultaneous equations. Before starting, you should make sure you are confident with rearranging formulae and solving linear equations (Factsheet 03 Algebraic Manipulation I). Example 2. Solve the equation 5x2 – 3x + 2 = (4 + x)(3 – x)

1. Quadratic equations A quadratic equation is any equation with x2 in it.

We must get this in the correct form. To do this, we must multiply out the brackets: 5x2 – 3x + 2 = 12 – 4x + 3x – x2

There are several ways to approach quadratic equations, but one way which always works is to use the quadratic equation formula. For the quadratic equation ax2 + bx + c = 0, the values of x are given by

Now we must get it all on one side of the equation: 5x2 – 3x + 2 – 12 + 4x – 3x + x2 = 0


b  b
4ac 2

x=

2a

Now collect the terms: 6x2 – 2x – 10 = 0

NB: If you are happy and confident with other methods, use them where applicable, but you should make sure you can use this method as well, since there are problems where it is the only way.

So a = 6, b = -2, c = -10 (You may have noticed that we could divide this whole equation by 2, to give 3x2 – x – 5 = 0. Using this with the formula would give the same answer)

Points to note
You must get the equation so that it has 0 on one side before trying to use this formula
You must collect terms if necessary, so that all the x2 terms are combined, all the x terms are combined and all the number terms are combined
It will help to avoid mistakes if you rearrange your equation in the correct order – that is, x2 terms, then x terms, then numbers
If you have x2 without a number in front of it, then a = 1
If you do not have a number term (eg 2x2 - 15x = 0), then c = 0
If you do not have an x term, then b is 0

So x

Using Your Calculator A few points to note:
In many calculators, if you type in –22, you will get the answer –4, not 4. This is because the calculator does the “square” before the “minus”. To avoid the problem, use brackets, as in the worked examples.
To avoid rounding errors, work out the square root (244 in the above example) and put it in the calculator memory.
Work out the number on the bottom of the fraction separately as well – as in the above example, where 2(6) has been replaced by 12.

So we substitute into the formula:


3  3 2
4(4)(-7)

2(4) The best strategy is to work out what is inside the square-root first: 32 – 4(4)(-7) = 121

Example 3. A body is moving in a straight line with constant acceleration 0.6ms-2. Its initial speed is 4 ms-1. After T seconds, it has travelled 9.5 m. Find the value of T.

Tips: 1. Be very careful with minus signs. Missing them out when doing the calculation is one of the commonest sources of error! 2. If you work out what is inside the square root and get a negative answer, you have gone wrong! Go back and check.

Use s = ut + ½ at2: 9.5 = 4T + ½ (0.6)T2 We must rearrange this to the standard form: 0.3T2 + 4T – 9.5 = 0


3  121

8 Now we must seperate out the
: So x =


3  121

8 So x = 1 or – 1.75

or

2(6)

2  244 12 = 1.47 (3 SF) or – 1.14 (3 SF)

This is already in the correct form a = 4, b = 3 and c = -7

So x =


(
2)  (
2)2
4(6)(-10)

=

Example 1. Solve the equation 4x2 + 3x – 7 = 0

x=

=

So T =


4  4 2
4(0.3)(-9.5)

2(0.3) = 2.06 s, -15.4 s (3SF) But time can’t be negative, so T = 2.06 s (3 SF)


3
121 8

1


   

Algebraic Manipulation 2 

Sometimes more complicated equations occur. However, exactly the same method can be used:

2. Simultaneous equations Simultaneous equations are pairs of equations, both of which have x and y (or a and b, or any other two variables) in them. One example is: x+y=3 2x + 3y = 7

Example 3. Solve the equations 4x2 + y2 = 5 6x +2y = 7

There are several good methods for solving this type of equation. Here we will use a method called substitution; this method is also applicable to more complicated examples.

1) Second equation: 6x + 2y = 7 3x + y = 3.5 (dividing by 2) y = 3.5 – 3x

The general method is:

2) 4x2 + (3.5 – 3x)2 = 5

1) Choose the easier looking equation and rearrange it to make y or x (it doesn’t matter which) the subject. 2) Use your answer to 1) to substitute into the other equation. You should end up with just one letter – x or y – not both. 3) Simplify the result of 2) and find the value of the letter you have. 4) Go back to your answer to 1) and use it to find the other letter.

3) 4x2 + ( 3.5 – 3x)(3.5 – 3x) = 5 4x2 + 12.25 – 21x + 9x2 = 5 13x2 – 21x + 7.25 = 0 This is a quadratic equation, so using the formula, we have:

Example 1. Solve the simultaneous equations x+y=3 2x + 3y = 7

x

=


(
21)  (
21)2
4(13)(7.25) 2  13

21  64 26 = 1.115 (4 SF), 0.5

= 1) The first equation looks easier. Making x the subject:x = 3 – y

4) y = 3.5 – x So if x = 1.115, y = -0.155 (3 SF) if x = 0.5, y = 3

2) We must replace the “x” in the second equation with “3 – y” 2(3 – y) + 3y = 7 Tip: Always take care to use brackets when you are substituting in.

Example 4. Solve the equations r=R+2 15 R= r 1) Already in the form r = R + 2

3) To simplify, we multiply out the brackets: 6 – 2y + 3y = 7 6+y=7 y=1 4) From 1), we know x = 3 – y But y = 1. So x = 3 – 1 = 2.

2) R =

3) R(R + 2) = 15 multiplying up R2 + 2R – 15 = 0

Example 2. Solve the simultaneous equations 8y  5x + 33 = 0 x + 4y = 1

R =

1) Second equation looks easier. It is easier to make x the subject: x = 1 – 4y

=

2) 8y – 5(1 – 4y) + 33 = 0


2  64 2

4) R = 3  r = 5 R = -5  r = 3

4) x = 1 – 4(-1) = 1 + 4 = 5

2. Solve the following simultaneous equations a) 5x + 4y = 21 x–y+3=0 b) 2x + 3y = 3 3x + 2y = 7 c) x2 + xy – x = 1 3x + y + 2 = 0


2  2 2
4(1)(-15) 21

R = 3, -5

3) 8y – 5 + 20y + 33 = 0 28y = -28  y = -1

Questions 1. Solve the following quadratic equations a) 3x2 – 8x + 2 = 0 b) 2x = 3 – x2 c) 0.3x2 – 4x + 2 = 0.7x(3 – 1.5x)

15 R2

Tip: If you have a graphical calculator, it may be able to solve basic simultaneous equations and quadratic equations automatically. This could save a lot of time! Answers 1. a) 2.39, 0.279 (3 SF) b) 1, -3 c) 4.16, 0.356 (3 SF) 2. a) x = 1, y = 4 b) x = 3, y = -1 c) x = -1, y =1 or x = -0.5, y = - 0.5 Acknowledgements: This Factsheet was researched and written by Cath Brown Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. They may be networked for use within the school. No part of these Factsheets may be reproduced, stored in a retrieval system or transmitted in any other form or by any other means without the prior permission of the publisher. ISSN 1351-5136

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