2 In this question you must show detailed reasoning.
Solve the equation \(3 x + 1 = 4 \sqrt { x }\).
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Question 2:
*(DR — detailed reasoning required)*
Main Method:
Answer Marks
Guidance
Answer Marks
Guidance
\(3(\sqrt{x})^2 + 1 = 4\sqrt{x}\) would be enough M1\ *Recognise as a quadratic in \(\sqrt{x}\); condone not \(= 0\)
\(3(\sqrt{x})^2 - 4\sqrt{x} + 1 = 0\) A1 All 3 terms on one side and \(= 0\)
\((3\sqrt{x}-1)(\sqrt{x}-1) = 0\) M1dep\ *Attempt to solve for \(\sqrt{x}\)
\(x = 1\) or \(x = \frac{1}{9}\) A1
Alternative Method:
Answer Marks
Guidance
Answer Marks
Guidance
\((3x+1)^2 = 16x\), \(\ 9x^2 + 6x + 1 = 16x\) M1\ *Square both sides; three terms from squaring bracket, at least 2 correct and \(= kx\) with \(k = 4\) or \(16\)
\(9x^2 - 10x + 1 = 0\) A1 All 3 terms on one side and \(= 0\)
\((9x-1)(x-1) = 0\) M1dep\ *Attempt to solve
\(x = 1\) or \(x = \frac{1}{9}\) A1
Total: [4]
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## Question 2:
*(DR — detailed reasoning required)*
**Main Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3(\sqrt{x})^2 + 1 = 4\sqrt{x}$ would be enough | **M1\*** | Recognise as a quadratic in $\sqrt{x}$; condone not $= 0$ |
| $3(\sqrt{x})^2 - 4\sqrt{x} + 1 = 0$ | **A1** | All 3 terms on one side and $= 0$ |
| $(3\sqrt{x}-1)(\sqrt{x}-1) = 0$ | **M1dep\*** | Attempt to solve for $\sqrt{x}$ |
| $x = 1$ or $x = \frac{1}{9}$ | **A1** | |
**Alternative Method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3x+1)^2 = 16x$, $\ 9x^2 + 6x + 1 = 16x$ | **M1\*** | Square both sides; three terms from squaring bracket, at least 2 correct and $= kx$ with $k = 4$ or $16$ |
| $9x^2 - 10x + 1 = 0$ | **A1** | All 3 terms on one side and $= 0$ |
| $(9x-1)(x-1) = 0$ | **M1dep\*** | Attempt to solve |
| $x = 1$ or $x = \frac{1}{9}$ | **A1** | |
**Total: [4]**
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2 In this question you must show detailed reasoning.
Solve the equation $3 x + 1 = 4 \sqrt { x }$.
\hfill \mbox{\textit{OCR PURE Q2 [4]}}