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Algebra

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Algebra

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  1. Algebra Chapter 1 LCH GH A Roche

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

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

  4. p.3 Simplify 1. 4x(3x2 + 5x + 6) – 2(10x2 + 12x) 4x(3x2 + 5x + 6) – 2(10x2 + 12x) = 12x3 + 20x2 + 24x – 20x2 - 24x = 12x3

  5. p.3 Simplify 2. (x + 2)2 + (x - 2)2 - 8 Expand the squares (x + 2)2 + (x - 2)2 - 8 = (x2 + 4x + 4) + (x2 - 4x + 4) - 8 = 2x2

  6. p.3 Simplify 3. (a + b)2 - (a - b)2 – 4ab Expand the squares (a + b)2 - (a - b)2 – 4ab = (a2 + 2ab + b2) - (a2 – 2ab + b2) – 4ab = a2 + 2ab + b2 - a2 + 2ab - b2 – 4ab = 0

  7. p.3 Simplify 4. (2a + b)2 – 4a(a + b) Expand (2a + b)2 – 4a(a + b) = (4a2 + 4ab + b2) - 4a2 – 4ab = 4a2 + 4ab + b2 - 4a2 – 4ab = b2

  8. p.3 Factorise 5. x2 + 3x x2 + 3x = x(x + 3) HCF Factorise 6. 3xy – 6y2 HCF 3xy – 6y2 = 3y(x - 2y)

  9. p.3 Factorise 7. a2b + ab2 a2b + ab2 = ab(a + b) HCF Factorise 8. 9x2 – 16y2 Difference of 2 squares 9x2 – 16y2 = (3x – 4y)(3x + 4y)

  10. p.3 Factorise 9. 121p2 – q2 Difference of 2 squares 121p2 – q2 = (11p – q)(11p + q) Factorise 10. 1 – 25a2 Difference of 2 squares 1 – 25a2 = (1 – 5a)(1 + 5a)

  11. p.3 Factorise 11. x2 – 2x - 8 -8 (1)(-8) (2)(-4) (4)(-2) (8)(-1) Quadratic factors Which factors add to -2? x2 – 2x - 8   = (x +2 )(x - 4 ) = (x )(x ) Factorise 12. 3x2 + 13x - 10 -10 (1)(-10) (2)(-5) (5)(-2) (10)(-1) 3x2 + 13x - 10 Quadratic Factors Check! +15x  = (3x )(x ) = (3x – 2)(x + 5) -2x

  12. p.3 Factorise 13. 6x2 - 11x + 3 6x2 (6x)(x) (3x)(2x) 6x2 - 11x + 3 Quadratic Factors Check! -9x   = ( )( ) = (3x )(2x ) = (3x – 1)(2x - 3) +3 (1)(3) (-1)(-3) -2x 

  13. p.7 Example 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 Expand the LHS • a(x2 + 2xb + b2) + c = RHS Observe that the LHS is a Quadratic Expression in x • ax2 + 2axb + ab2 + c = RHS • Equate coefficients of like terms • (a)x2 + (2ab)x + (ab2 + c) = 2x2 + 12x + 23 • 2ab = 12 • a = 2 • ab2 +c = 23 • 2(2)b = 12 • (2)(3)2 +c = 23 • 4b = 12 • 18 +c = 23 • b = 3 • c = 5

  14. 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 Expand the LHS • ax(x2 –px + 1) +k(x2 –px + 1) = RHS • 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 • k = c • a = a • k - ap = 0 • a – kp = b • c - ap = 0 • a – c(c/a) = b • a – c2 /a = b • c = ap • a – b = c2 /a • p = c/a • a(a – b) = c2

  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 Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 (i) 32 = (16 x 2) • = 162 • = 42 (ii) 45 = (9 x 5) • = 35 • (iii) 75 =(25 x 3) • = 53

  16. 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 (iv) (v)

  17. 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. Divide by the largest square number: 1 4 9 16 25 36 49 64 81 100 121 144 169 (i) 18 + 50 - 8 = (9 x 2) + (25 x 2) - (4 x 2) • = 32 + 52 – 22 • = 62 (ii) 20 - 5 + 45 • = (4 x 5) - 5 + (9 x 5) = 25 - 5 + 35 • = 45

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

  20. Conjugate Surds P.10 When a compound surd is multiplied by its conjugate the result is a rational number. • Conjugate 2 Compound Surd • Conjugate 1 • a - b • - a + b a + b • a - b • a + b • -a - b We use this ‘trick’ to solve fractions with compound denominators • a + b • a - b • - a + b

  21. p.10 Example Show that Multiply top and bottom by conjugate of denominator Note that the bottom is difference of 2 squares 1 – 3 Q.E.D.

  22. 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 Get rid of fractions by Multiplying x + 1 – y + 3 = 4 2 3 x + y – 3 = 1 • 2 2 • (6)(x + 1)– (6)(y + 3) = (6)4 • 2 3 2x + 2(y – 3) = 2(1) • (3)(x + 1)– (2)(y + 3) = 24 • 2 2 3x + 3 – 2y - 6 = 24 2x + y – 3 = 1 • 3x– 2y = 27 • 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 1 2 Label the equations 1, 2 & 3 • Eliminate z from 2 equations 3 5x – 3y +2z = 19 3x + 2y – 3z = -5 2 3 -2x - 4y -2z = -6 x -2 3x + 6y + 3z = 9 x 3 1 1 5 • 3x - 7y = 13 • 6x + 8y = 4 4 • Now solve simultaneous equations 4 & 5 in the usual way We find that: x = 2 y = -1 x + 2y + z = 3 • Sub these values into equation 1 • (2) + 2(-1) + z = 3 • z = 3

  25. 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 • 3x + 2y - z = -3 • 5x – 3y +2z = 3 • 5x + 3z = 14 1 • e.g. Solve 2 3 • 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