June 2016 Paper 2 Q15
15
(a) Solve \(\quad \dfrac{2}{5}\sqrt{x} = 1\) [2 marks]
(b) Solve \(\quad x^3 = 5x^2\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt{x} = \dfrac{5}{2}\) or \(\sqrt{x} = 2.5\) or \(\sqrt{x} = \dfrac{1}{0.4}\) or \(\dfrac{4}{25}x = 1\) or \(4x = 25\) | M1 | oe Must have \(\sqrt{x} = \ldots\) or have eliminated \(\sqrt{\phantom{x}}\) with no errors |
| \(\dfrac{25}{4}\) or 6.25 | A1 | oe |
Additional guidance
| \(2\sqrt{x} = 5\) (no further correct work) | M0 |
| Allow unprocessed values for M1 eg \(\dfrac{4}{25}x = 1^2\) or \(\left(\dfrac{2}{5}\right)^2x = 1\) | M1 |
| \(\sqrt{x}\) may be seen as \(x^{\frac{1}{2}}\) or \(x^{0.5}\) |
| Answer | Mark | Comments |
|---|---|---|
| \(x^2(x - 5)\ (= 0)\) or \(x^2(5 - x)\ (= 0)\) or \((x =)\ 0\) or \((x =)\ 5\) | M1 | oe factorisation eg1 \((x^2 - 0)(x - 5)\) eg2 \(x(x^2 - 5x)\) |
| 0 and 5 with no other solutions | A1 |
Additional guidance
| For A1, the word ‘and’ is not needed If their answer has both 0 and 5 with no other solutions award M1 A1 eg 0, 5 with no other solutions | M1 A1 |
| 0, 5, \(-5\) | M1 A0 |
| Either or both solutions seen embedded | M1 A0 |