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The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence HW1 – Grade C Trial and Improvement 1. (a) x2 + 4x Factorise ……………...........………………………….....……………………………………. Answer …………………………………. (1) (b) Solve the inequality 7y < 3y + 6 ……………...........………………………….....……………………………………. ……………...........………………………….....……………………………………. Answer …………………………………. (2) (c) Make r the subject of the formula p = 3 + 2r ……………...........………………………….....……………………………………. ……………...........………………………….....……………………………………. ……………...........………………………….....……………………………………. Answer r = …………………………………. (2) (Total 5 marks) 2. Laura is using trial and improvement to find a solution to the equation x3+ 2x = 60 The table shows her first try. Continue the table to find a solution to the equation. x Comment x3 + 2x 3 33 too small Give your answer correct to 1 decimal place. Answer x = .................................................... (Total 4 marks) The Robert Smyth School 1 The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence 3. Gary is using trial and improvement to find a solution to the equation x3– 5x = 56. This table shows his first two trials. x x3 – 5x Comment 4 44 Too small 5 100 Too big Continue the table to find a solution to the equation. Give your answer to 1 decimal place. Answer x = ................................................................ (Total 3 marks) 4. Parveen is using trial and improvement to find a solution to the equation x3 + 7x = 30 This table shows her first two trials. x x3 + 7x Comment 2 22 Too small 3 48 Too big Continue the table to find a solution to the equation. Give your answer to 1 decimal place. Answer ............................................................. (Total 3 marks) The Robert Smyth School 2 The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence 5. Use trial and improvement to find a solution to the equation x3 – x = 21 Give your answer to one decimal place. You must show your working. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ Answer x = ………………………………. (Total 4 marks) 6. A solution of the equation x3 – 8x = 110 lies between x = 5 and x = 6. Use trial and improvement to find this solution. Give your answer to one decimal place. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... Answer x = .................................................................. (Total 3 marks) The Robert Smyth School 3 The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence 7. Kate and Lee are working out this question. A solution of the equation x3 + x = 700 lies between 8 and 9. Use trial and improvement to find this solution, correct to one decimal place. Kate’s answer is 8.8 Lee’s answer is 8.9 Which answer is correct? You must show all your working. ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... Answer x = ...................................................... (Total 3 marks) The Robert Smyth School 4 The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence 1. (a) x(x + 4) (b) 4y < 6 B1 M1 y = 1.5 oe scores M1A0 y < 1.5 A1 oe (c) p – 3 = 2r or p/2 = 3/2 + r or (p + 3)/2 or (3 – p)/2 M1 (p – 3)/2 or p/2 – 3/2 oe A1 [5] 2. 3.7 M1,M1,A1,A1 M1 for trying value bigger than 3 M1 for trying 1d.p value between 3 and 4 M1 for sandwiching between 3.7 and 3.8 A1 for testing 3.75 (or other apt 2 dp value) and stating answer [4] 3. Trial for 4 < x < 5 B1 Correctly evaluated to at least nearest whole number 2 trials for 4.2 x 4.3 that ‘bracket’ 56 Trials must be correct to or truncated to at least 1 dp B1 Trial at 4.25 & answer 4.3 B1 [3] 4. Trial between 2 and 3 B1 Trial between 2.3 and 2.4 inclusive that “bracket” the answer B1 Trial at 2.35 or 2.36 or 2.37 and 2.4 stated as answer DB1 In this question final answer on its own will not get any marks Working must be seen. All trials must be correctly evaluated either rounded or truncated to a degree of accuracy that allows comparison. [3] 5. Trial above 2.8796 M1 2 gives 6, 3 gives 24 Trial below 2.8796 M1 2.9 gives 21.489, all values to at least l dp rounded or truncated Testing a value that justifies 2.9 as answer 2.5 13.125, 2.6 14.976, 2.7 16.983 2.8 19.152, 2.85 gives 20.299 x = 2.9 The Robert Smyth School DM1 A1 5 The Robert Smyth School Mathematics Faculty Topic 9 Solving Equations 2 Innovation & excellence Dep on any M mark [4] 6. Trial between 5 and 6 B1 Trials at 5.3 and 5.4 or better that bracket 110 5.3 106(.47...) 5.4 114(.26...) B1 Trial at 5.35 and answer 5.3 5.35 110.3(3...) B1 [3] 7. Trial at 8.8 or 8.9 (or between) 690(.272) or 713(.869) B1 Trials at 8.85 (702) and 8.8 or any 2 trials for 8.8 ≤ x ≤ 8.85 which bracket 700 Note: These 2 trials score first B1 also B1 8.8 or Kate B1 dep Dependent on second B1 Alternative method: Trial at 8.85 M1 702 A1 8.8 A1 [3] The Robert Smyth School 6