| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Substitution to solve disguised quadratic |
| Difficulty | Moderate -0.8 This is a straightforward substitution question requiring students to recognize that x = u² when u = √x, transform to a quadratic in u, solve using standard methods (factorization or formula), then back-substitute. While it requires multiple steps, the technique is routine and commonly practiced in AS-level courses with no conceptual challenges beyond applying a given substitution. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(u = \sqrt{x}\): \(6x + 7\sqrt{x} - 20 = 0 \Rightarrow 6u^2 + 7u - 20 = 0 \Rightarrow (3u-4)(2u+5) = 0\) | M1A1 | M1: valid method (substitution, factorisation attempt, formula, completing square); A1: correct factorisation or equivalent |
| Attempts \(\sqrt{x} = \text{"}\frac{4}{3}\text{"},\text{"}-\frac{5}{2}\text{"} \Rightarrow x = \ldots\) | M1 | Squares their value(s) of \(u\) to get \(x\) |
| \(x = \frac{16}{9}\) only | A1 cso | \(x = \frac{25}{4}\) must be discarded; allow "incorrect" \(x = -\frac{16}{9}\) or \(x = -\frac{25}{4}\) seen and discarded; ignore reason given for rejecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(7\sqrt{x} = 20 - 6x \Rightarrow 49x = (20-6x)^2 \Rightarrow 36x^2 - 289x + 400 = 0\) | M1, A1 | |
| \((9x-16)(4x-25) = 0\) | M1 | Must factorise with \(ab=36, cd=400\) |
| \(x = \frac{16}{9}\) only | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((3\sqrt{x}-4)(2\sqrt{x}+5) = 0\) | M1, A1 | \(ab=6, cd=20\) |
| Attempts \(\sqrt{x} = \text{"}\frac{4}{3}\text{"},\text{"}-\frac{5}{2}\text{"} \Rightarrow x = \ldots\) | M1 | |
| \(x = \frac{16}{9}\) only | A1 cso |
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $u = \sqrt{x}$: $6x + 7\sqrt{x} - 20 = 0 \Rightarrow 6u^2 + 7u - 20 = 0 \Rightarrow (3u-4)(2u+5) = 0$ | M1A1 | M1: valid method (substitution, factorisation attempt, formula, completing square); A1: correct factorisation or equivalent |
| Attempts $\sqrt{x} = \text{"}\frac{4}{3}\text{"},\text{"}-\frac{5}{2}\text{"} \Rightarrow x = \ldots$ | M1 | Squares their value(s) of $u$ to get $x$ |
| $x = \frac{16}{9}$ only | A1 cso | $x = \frac{25}{4}$ must be discarded; allow "incorrect" $x = -\frac{16}{9}$ or $x = -\frac{25}{4}$ seen and discarded; ignore reason given for rejecting |
**Alt 1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $7\sqrt{x} = 20 - 6x \Rightarrow 49x = (20-6x)^2 \Rightarrow 36x^2 - 289x + 400 = 0$ | M1, A1 | |
| $(9x-16)(4x-25) = 0$ | M1 | Must factorise with $ab=36, cd=400$ |
| $x = \frac{16}{9}$ only | A1 cso | |
**Alt 2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(3\sqrt{x}-4)(2\sqrt{x}+5) = 0$ | M1, A1 | $ab=6, cd=20$ |
| Attempts $\sqrt{x} = \text{"}\frac{4}{3}\text{"},\text{"}-\frac{5}{2}\text{"} \Rightarrow x = \ldots$ | M1 | |
| $x = \frac{16}{9}$ only | A1 cso | |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
Using the substitution $u = \sqrt { x }$ or otherwise, solve
$$6 x + 7 \sqrt { x } - 20 = 0$$
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q2 [4]}}