Intermediate November 2020 Unit 2 Q12
12. A solution of the equation
\[x^3 - 5x - 350 = 0\]lies between 7·2 and 7·3.
Use the method of trial and improvement to find this solution correct to 2 decimal places.
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| One correct evaluation 7·2 ≤ x ≤ 7·3 | B1 | Correct evaluation regarded as enough to identify if negative or positive. If evaluations not seen accept ‘too high’ or ‘too low’. Look out for equating x3 – 5x = 350 |
| 2 correct evaluations 7·275 ≤ x ≤ 7·295, one < 0, one > 0. | B1 | |
| 2 correct evaluations 7·275 ≤ x ≤ 7·285, one < 0, one > 0. | M1 | |
| x = 7·28 | A1 |
Table of values
| x | x3 – 5x – 350 | ||
|---|---|---|---|
| 7·2 | –12·75(2) | ||
| 7·21 | –11(·2.. ) | ||
| 7·22 | – 9(·7…) | ||
| 7·23 | – 8(·2…) | ||
| 7·24 | – 6(·6…) | ||
| 7·25 | –5(·1...) | ||
| 7·26 | – 3(·6…) | 7·275 | −1(·3….) |
| 7·27 | – 2(·1…) | 7·284 | 0(·04..) |
| 7·28 | −0·5(7..) | 7·285 | 0·1(9..) |
| 7·29 | 0·9(7..) | 7·295 | 1(·7….) |
| 7·3 | 2·5(17) |
In the original this table is printed in the Comments column; the bold entries are bold in the original.