Higher June 2019 Paper 1 Q17
17 Toby is forming and solving equations.
(a)
| The product of half of a number and three more than the number is the same as the square of the number |
Toby uses \(y\) to represent the number.
Write an equation that Toby could form. [2 marks]
(b) Toby forms another equation.\[x = \frac{9}{8x}\]
He wants to work out the values of \(x\).
Here is his working.
| \(x = \dfrac{9}{8x}\) \(8x^2 = 9\) \(8x = 3 \quad\text{or}\quad 8x = -3\) \(x = \dfrac{3}{8} \quad\text{or}\quad x = -\dfrac{3}{8}\) |
What error has he made in his working? [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(y^2 = \dfrac{1}{2}y(y + 3)\) | B2 | oe equation eg \(2y^2 = y^2 + 3y\) or \(y^2 = 3y\) or \(y = 0\) or \(y = 3\) or \(y = 0\) or 3 B1 \(\dfrac{1}{2}y(y + 3)\) oe expression or an otherwise correct equation using a different unknown or combination of unknowns |
Additional guidance
| Allow multiplication signs eg \(y \times y = \dfrac{y}{2} \times (y + 3)\) | B2 |
| \(y^2 = \dfrac{1}{2}y(y + 3)\) followed by incorrect simplification or attempt to solve the equation | B2 |
| \(y^2 = \dfrac{1}{2}y + y + 3\) | B0 |
| 3 only or 0 only or 0 and 3 only | B0 |
| Do not allow missing or partially missing brackets unless recovered eg1 \(y^2 = \dfrac{1}{2}y \times y + 3\) without correct equation seen eg2 \(y^2 = \dfrac{1}{2}y(y + 3\) without correct equation seen | B0 B0 |
| Answer | Mark | Comments |
|---|---|---|
| Correct comment or shows correct working | B1 | eg1 he hasn’t square rooted (correctly) eg2 it should be \(\sqrt{8}\,x = 3\) eg3 he should have divided (by 8) before square rooting |
Additional guidance
| \(\sqrt{8}\) may be given as \(2\sqrt{2}\) | |
| Comment that he shouldn’t have a negative answer | B0 |
| Mathematically incorrect statement | B0 |
| Correct comment and an incorrect comment | B0 |
Example responses
| He has taken it as \((8x)^2\) | B1 |
| He has divided \(8x^2\) by \(x\) (instead of square rooting) and square rooted the 9 | B1 |
| He \(\sqrt{\phantom{x}}\) first when supposed to divide it by 8 | B1 |
| He didn’t divide 9 by 8 to get \(x^2\) | B1 |
| At the start he took the 8 over when you want \(\sqrt{\dfrac{9}{8}}\) | B1 |
| Toby should have got \(\pm\sqrt{\dfrac{9}{8}}\) | B1 |
| He should have divided by 8 | B0 |
| Toby didn’t square root \(8x\) | B0 |
| He hasn’t square rooted the \(8x^2\) to leave \(x\) on its own | B0 |
| He hasn’t square rooted the other side to just get \(x\) | B0 |
| Didn’t divide by 8 | B0 |
| He should have divided by \(8x\) | B0 |
| He found the square root of 9 but didn’t write \(\sqrt{8x} = 9\) | B0 |