Algebra

Chapter 1LCH GH

A Roche

Simplify (i)

=

=

=

multiply each part by x

factorise the top

Divide top & bottom by (x-3)

p.3

+Simplify (ii)

+ Multiply second part above and below by -1So that both denominators are the same

Factorise the top

p.3

p.3Simplify 1. 4x(3x2 + 5x + 6) – 2(10x2 + 12x)

= 12x3 + 20x2 + 24x – 20x2 - 24x= 12x3

4x(3x2 + 5x + 6) – 2(10x2 + 12x)

p.3Simplify 2. (x + 2)2 + (x - 2)2 - 8

= (x2 + 4x + 4) + (x2 - 4x + 4) - 8= 2x2

(x + 2)2 + (x - 2)2 - 8 Expand the squares

p.3Simplify 3. (a + b)2 - (a - b)2 – 4ab

= (a2 + 2ab + b2) - (a2 – 2ab + b2) – 4ab= 0

Expand the squares (a + b)2 - (a - b)2 – 4ab= a2 + 2ab + b2 - a2 + 2ab - b2 – 4ab

p.3Simplify 4. (2a + b)2 – 4a(a + b)

= (4a2 + 4ab + b2) - 4a2 – 4ab= b2

Expand (2a + b)2 – 4a(a + b)= 4a2 + 4ab + b2 - 4a2 – 4ab

p.3Factorise5. x2 + 3x= x(x + 3) x2 + 3x HCF

Factorise6. 3xy – 6y2

3xy – 6y2 HCF= 3y(x - 2y)

p.3Factorise7. a2b + ab2

= ab(a + b) a2b + ab2 HCF

Factorise8. 9x2 – 16y2

9x2 – 16y2 Difference of 2 squares

= (3x – 4y)(3x + 4y)

p.3Factorise9. 121p2 – q2

= (11p – q)(11p + q) 121p2 – q2 Difference of 2 squares

Factorise10. 1 – 25a2

1 – 25a2 Difference of 2 squares

= (1 – 5a)(1 + 5a)

p.3Factorise11. x2 – 2x - 8

= (x )(x ) x2 – 2x - 8 Quadratic factors

Factorise12. 3x2 + 13x - 103x2 + 13x - 10 Quadratic

Factors= (3x – 2)(x + 5)Check!+15x

-2x

-8(1)(-8)(2)(-4)(4)(-2)(8)(-1)

-10(1)(-10)(2)(-5)(5)(-2)

(10)(-1)

Which factors add to -2?= (x +2 )(x - 4 )

= (3x )(x )

p.3Factorise13. 6x2 - 11x + 36x2 - 11x + 3 Quadratic

Factors= (3x – 1)(2x - 3)Check!-9x

-2x+3

(1)(3)(-1)(-3)

= ( )( ) 6x2

(6x)(x)(3x)(2x)

= (3x )(2x )

p.7 Example

(i) If a(x + b)2 + c = 2x2 + 12x + 23, for all x, find the value of a, of b and of c.

a(x + b)2 + c = 2x2 + 12x + 23

a(x2 + 2xb + b2) + c = RHS

Expand the LHS

ax2 + 2axb + ab2 + c = RHSObserve that the LHS is a Quadratic Expression in x

(a)x2 + (2ab)x + (ab2 + c) = 2x2 + 12x + 23

Equate coefficients of like terms

a = 2 2ab = 12

2(2)b = 12

4b = 12

b = 3

ab2 +c = 23

(2)(3)2 +c = 23

18 +c = 23

c = 5

p.7 Example

(ii) If (ax + k)(x2 – px +1) = ax3 + bx + c, for all x, show that c2 = a(a – b).

(ax + k)(x2 – px + 1) = ax3 + bx + c

ax(x2 –px + 1) +k(x2 –px + 1) = RHS

Expand the LHS

ax3 - apx2 + ax + kx2 –kpx + k = RHS

(a)x3 + (-ap + k)x2 + (a - kp)x + k = ax3 + 0x2 + bx + c

Equate coefficients of like terms

a = a k - ap = 0 a – kp = b k = c

c - ap = 0

c = ap

p = c/a

a – c(c/a) = b

a – c2 /a = b

a – b = c2 /a

a(a – b) = c2

p.9 Example

Write out each of the following in the form ab, where b is prime:(i) 32 (ii) 45 (iii) 75

Divide by the largest square number:

149

162536496481

100121144169

(i) 32

(iii) 75

(ii) 45

= (16 x 2) = 162 = 42

= (9 x 5) = 35

=(25 x 3) = 53

p.9 Example

Express in the form , a, b N :

(iv) (v)

Divide by the largest square number:

149

162536496481

100121144169

(iv)

(v)

p.9 Example

(vi) Express in the form k2.

(vi)

p.9 Example

(i) Express 18 + 50 - 8 in the form ab, where b is prime.

(ii) 20 - 5 + 45 = k5; find the value of k.Divide by the largest

square number:149

162536496481

100121144169

(i) 18 + 50 - 8

(ii) 20 - 5 + 45

= (9 x 2) + (25 x 2) - (4 x 2)

= 62

= 25 - 5 + 35

= 45

= 32 + 52 – 22

= (4 x 5) - 5 + (9 x 5)

Examples of Compound Surds

P.10

a- b

a- b

a + b 1 + 5

3- 24

13- 7

Conjugate Surds

P.10

a - b

a + b

a + b

Compound Surd Conjugate 1 Conjugate 2

a - b - a + b

a + b

a - b - a + b

-a - b

When a compound surd is multiplied by its conjugate the result is a rational number.

We use this ‘trick’ to solve fractions with compound denominators

p.10 Example

Show that

Multiply top and bottom by conjugate of denominator

Note that the bottom is difference of 2 squares

Q.E.D.

1 – 3

p.11

Solving Simultaneous Equations

For complicated simultaneous equations we use the substitution-elimination method

p.12 ExampleSolve for x and y the simultaneous equations:

x + 1 – y + 3 = 4, x + y – 3 = 1 2 3 2 2

Get rid of fractions by Multiplying

x + 1 – y + 3 = 4 2 3

(6)(x + 1)– (6)(y + 3) = (6)4 2 3

x + y – 3 = 12 2

(3)(x + 1)– (2)(y + 3) = 24

3x + 3 – 2y - 6 = 24

3x– 2y = 27

2x + 2(y – 3) = 2(1) 2 2

2x + y – 3 = 1

2x + y = 4

Now solve these simultaneous equations in the normal way

x = 5 and y = -6

p.12-13 Example

Solve for x, y and z: x + 2y + z = 35x – 3y +2z = 193x + 2y – 3z = -5

Label the equations 1, 2 & 3

2

1

3

21

Eliminate z from 2 equations

-2x - 4y -2z = -65x – 3y +2z = 19

x -2

3x - 7y = 13 4

3x + 2y – 3z = -53x + 6y + 3z = 9

3

1 x 3

6x + 8y = 4 5

Now solve simultaneous equations 4 & 5 in the usual way

Sub these values into equation 1

x + 2y + z = 3

(2) + 2(-1) + z = 3

z = 3

x = 2

y = -1

We find that:

p.13

Note:

If one equation contains only 2 variables then the other 2 equations are used to obtain a second equation with the same two variables

Here, from equation 1 and 2, y should be eliminated to obtain an equation in x and z, which should then be used with equation 3

3x + 2y - z = -35x – 3y +2z = 35x + 3z = 14

e.g. Solve

3

2

1