1. Algebra Chapter 1 LCH GH A Roche

2. Simplify (i) = = = multiply each part by x factorise the top Divide top & bottom by (x-3) p.3

3. + Simplify (ii) + Multiply second part above and below by -1 So that both denominators are the same Multiply second part above and below by -1 So that both denominators are the same Factorise the top p.3

4. Simplify 1. 4x(3x 2 + 5x + 6) – 2(10x 2 + 12x) Simplify 1. 4x(3x 2 + 5x + 6) – 2(10x 2 + 12x) = 12x 3 + 20x 2 + 24x – 20x 2 - 24x = 12x 3 4x(3x 2 + 5x + 6) – 2(10x 2 + 12x)

5. p.3 Simplify 2. (x + 2) 2 + (x - 2) 2 - 8 Simplify 2. (x + 2) 2 + (x - 2) 2 - 8 = (x 2 + 4x + 4) + (x 2 - 4x + 4) - 8 = 2x 2 (x + 2) 2 + (x - 2) 2 - 8 Expand the squares

6. p.3 Simplify 3. (a + b) 2 - (a - b) 2 – 4ab Simplify 3. (a + b) 2 - (a - b) 2 – 4ab = (a 2 + 2ab + b 2 ) - (a 2 – 2ab + b 2 ) – 4ab = 0 Expand the squares (a + b) 2 - (a - b) 2 – 4ab = a 2 + 2ab + b 2 - a 2 + 2ab - b 2 – 4ab

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

8. p.3 Factorise 5. x 2 + 3x Factorise 5. x 2 + 3x = x(x + 3) x 2 + 3x HCF Factorise 6. 3xy – 6y 2 Factorise 6. 3xy – 6y 2 3xy – 6y 2 HCF = 3y(x - 2y)

9. p.3 Factorise 7. a 2 b + ab 2 Factorise 7. a 2 b + ab 2 = ab(a + b) a 2 b + ab 2 HCF Factorise 8. 9x 2 – 16y 2 Factorise 8. 9x 2 – 16y 2 9x 2 – 16y 2 Difference of 2 squares = (3x – 4y)(3x + 4y)

10. p.3 Factorise 9. 121p 2 – q 2 Factorise 9. 121p 2 – q 2 = (11p – q)(11p + q) 121p 2 – q 2 Difference of 2 squares Factorise 10. 1 – 25a 2 Factorise 10. 1 – 25a 2 1 – 25a 2 Difference of 2 squares = (1 – 5a)(1 + 5a)

11. p.3 Factorise 11. x 2 – 2x - 8 Factorise 11. x 2 – 2x - 8 = (x )(x ) x 2 – 2x - 8 Quadratic factors Factorise 12. 3x 2 + 13x - 10 Factorise 12. 3x 2 + 13x - 10 3x 2 + 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 )

12. p.3 Factorise 13. 6x 2 - 11x + 3 Factorise 13. 6x 2 - 11x + 3 6x 2 - 11x + 3 Quadratic Factors = (3x – 1)(2x - 3) Check! -9x -2x +3 (1)(3) (-1)(-3) = ( )( ) 6x 2 (6x)(x) (3x)(2x) = (3x )(2x )

13. p.7 Example (i)If a(x + b) 2 + c = 2x 2 + 12x + 23, for all x, find the value of a, of b and of c. p.7 Example (i)If a(x + b) 2 + c = 2x 2 + 12x + 23, for all x, find the value of a, of b and of c. a(x + b) 2 + c = 2x 2 + 12x + 23 a(x 2 + 2xb + b 2 ) + c = RHS Expand the LHS ax 2 + 2axb + ab 2 + c = RHS Observe that the LHS is a Quadratic Expression in x (a)x 2 + (2ab)x + (ab 2 + c) = 2x 2 + 12x + 23 Equate coefficients of like terms a = 2 2ab = 12 2(2)b = 12 4b = 12 b = 3 ab 2 +c = 23 (2)(3) 2 +c = 23 18 +c = 23 c = 5

14. p.7 Example (ii)If (ax + k)(x 2 – px +1) = ax 3 + bx + c, for all x, show that c 2 = a(a – b). p.7 Example (ii)If (ax + k)(x 2 – px +1) = ax 3 + bx + c, for all x, show that c 2 = a(a – b). (ax + k)(x 2 – px + 1) = ax 3 + bx + c ax(x 2 –px + 1) +k(x 2 –px + 1) = RHS Expand the LHS ax 3 - apx 2 + ax + kx 2 –kpx + k = RHS (a)x 3 + (-ap + k)x 2 + (a - kp)x + k = ax 3 + 0x 2 + 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 – c 2 /a = b a – b = c 2 /a a(a – b) = c 2

15. p.9 Example Write out each of the following in the form a  b, where b is prime: (i)  32(ii)  45(iii)  75 p.9 Example Write out each of the following in the form a  b, where b is prime: (i)  32(ii)  45(iii)  75 Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 (i)  32 (iii)  75 (ii)  45 =  (16 x 2)=  16  2= 4  2 =  (9 x 5)= 3  5 =  (25 x 3)= 5  3

16. p.9 Example Express in the form, a, b  N : (iv)(v) p.9 Example Express in the form, a, b  N : (iv)(v) Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 (iv) (v)

17. p.9 Example (vi) Express in the form k  2. p.9 Example (vi) Express in the form k  2. (vi)

18. p.9 Example (i)Express  18 +  50 -  8 in the form a  b, where b is prime. (ii)  20 -  5 +  45 = k  5; find the value of k. p.9 Example (i)Express  18 +  50 -  8 in the form a  b, where b is prime. (ii)  20 -  5 +  45 = k  5; find the value of k. Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 (i)  18 +  50 -  8 (ii)  20 -  5 +  45 =  (9 x 2) +  (25 x 2) -  (4 x 2) = 6  2 = 2  5 -  5 + 3  5 = 4  5 = 3  2 + 5  2 – 2  2 =  (4 x 5) -  5 +  (9 x 5)

19. Examples of Compound Surds Examples of Compound Surds P.10  a-  b a +  b 1 +  5  3- 2  4  13- 7

20. Conjugate Surds Conjugate Surds P.10  a -  b  a +  b a +  b Compound SurdConjugate 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

21. p.10 Example Show that 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

22. p.11 Solving Simultaneous Equations p.11 Solving Simultaneous Equations For complicated simultaneous equations we use the substitution-elimination method

23. p.12 Example Solve for x and y the simultaneous equations: x + 1 – y + 3 = 4,x + y – 3 = 1 2 3 2 2 p.12 Example Solve 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 = 1 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

24. p.12-13 Example Solve for x, y and z: x + 2y + z = 3 5x – 3y +2z = 19 3x + 2y – 3z = -5 p.12-13 Example Solve for x, y and z: x + 2y + z = 3 5x – 3y +2z = 19 3x + 2y – 3z = -5 Label the equations 1, 2 & 3 2 1 3 2 1 Eliminate z from 2 equations -2x - 4y -2z = -6 5x – 3y +2z = 19 x -2 3x - 7y = 13 4 3x + 2y – 3z = -5 3x + 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:

25. p.13 Note: 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 = -3 5x – 3y +2z = 3 5x + 3z = 14 e.g. Solve 3 2 1