Higher June 2023 Unit 2 Q15
15. Use the quadratic formula to solve the equation \(\dfrac{1}{x - 2} + \dfrac{1}{3x - 7} = 1\).
Give your answers correct to 2 decimal places.
You must show all your working. [6]
From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| A correct common denominator may be shown throughout for first 3 marks. Also, look out for alternative, correct methods leading to the same quadratic equated to zero. | ||
| \(3x - 7 + x - 2 = (x - 2)(3x - 7)\) oe | B1 | For multiplying throughout the numerator of all terms by the common denominator. |
| Sight of \(3x^2 - 7x - 6x + 14\) | B1 | Or equivalent. May be seen in the denominator. |
| \(3x^2 - 17x + 23 = 0\) OR \(-3x^2 + 17x - 23 = 0\) | B1 | '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 3, b = -17\), \(c = 23\) used in the quadratic formula. FT from B1B0 from one error only. |
| \(x = \dfrac{-(-17) \pm \sqrt{(-17)^2 - 4 \times 3 \times 23}}{2 \times 3}\) | M1 | This substitution into the formula must be seen for M1, otherwise award M0A0A0. FT ‘their derived quadratic equation equated to 0’, (but not \(3x^2 - 13x + 14 = 0\)), provided of equivalent difficulty (\(a\), \(b\) and \(c\) must be non-zero). Allow one slip in substitution for M1 only, but must be correct formula. |
| \(x = \dfrac{17 \pm \sqrt{13}}{6}\) | A1 | Can be implied from at least one correct value of \(x\) evaluated. |
| \(x\) = 3·43 with \(x\) = 2·23 | A1 | CAO for their quadratic equation. Answers must be given to 2 decimal places. On FT solutions must require rounding. |