Edexcel C1 2014 June — Question 5 4 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2014
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeSolve equations with surds
DifficultyModerate -0.8 This is a straightforward surd manipulation question requiring simplification of √8 and rationalization of the denominator, followed by solving a linear equation. It's easier than average as it involves only routine algebraic techniques with no problem-solving insight needed, though it requires more steps than the most basic index law recall questions.
Spec1.02b Surds: manipulation and rationalising denominators

5. Solve the equation $$10 + x \sqrt { 8 } = \frac { 6 x } { \sqrt { 2 } }$$ Give your answer in the form \(a \sqrt { } b\) where \(a\) and \(b\) are integers.

Question 5:
Method 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x\sqrt{8} + 10 = \frac{6x}{\sqrt{2}}\) Starting equation
\(\times\sqrt{2} \Rightarrow x\sqrt{16} + 10\sqrt{2} = 6x\)M1, A1 For multiplying both sides by \(\sqrt{2}\) – allow a slip e.g. \(\sqrt{2}x\sqrt{8}+10=\frac{6x}{\sqrt{2}}\times\sqrt{2}\) or \(\sqrt{2}\times10+x\sqrt{8}=\frac{6x}{\sqrt{2}}\times\sqrt{2}\), where one term has an error. NB \(x\sqrt{8}+10=6x\sqrt{2}\) is M0
\(4x + 10\sqrt{2} = 6x \Rightarrow 2x = 10\sqrt{2}\), \(x = 5\sqrt{2}\), \(a=5\), \(b=2\)M1, A1 A1: correct equation in \(x\) with no fractional terms e.g. \(x\sqrt{16}+10\sqrt{2}=6x\). M1: attempt to solve linear equation producing answer of form \(a\sqrt{2}\) or \(a\sqrt{50}\). A1: \(5\sqrt{2}\) oe (accept \(1\sqrt{50}\))
Method 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2\sqrt{2}x + 10 = 3\sqrt{2}x\)M1, A1 M1: writing \(\sqrt{8}\) as \(2\sqrt{2}\) or \(\frac{6}{\sqrt{2}}\) as \(3\sqrt{2}\). A1: correct equation with no fractional terms e.g. \(2\sqrt{2}x+10=3\sqrt{2}x\) or \(x\sqrt{8}+10=3\sqrt{2}x\)
\(\sqrt{2}x = 10 \Rightarrow x = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}\) oeM1, A1 M1: attempt to solve producing answer of form \(a\sqrt{2}\) or \(a\sqrt{50}\). A1: \(5\sqrt{2}\) oe, accept \(1\sqrt{50}\)
## Question 5:

**Method 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\sqrt{8} + 10 = \frac{6x}{\sqrt{2}}$ | | Starting equation |
| $\times\sqrt{2} \Rightarrow x\sqrt{16} + 10\sqrt{2} = 6x$ | M1, A1 | For multiplying both sides by $\sqrt{2}$ – allow a slip e.g. $\sqrt{2}x\sqrt{8}+10=\frac{6x}{\sqrt{2}}\times\sqrt{2}$ or $\sqrt{2}\times10+x\sqrt{8}=\frac{6x}{\sqrt{2}}\times\sqrt{2}$, where one term has an error. NB $x\sqrt{8}+10=6x\sqrt{2}$ is M0 |
| $4x + 10\sqrt{2} = 6x \Rightarrow 2x = 10\sqrt{2}$, $x = 5\sqrt{2}$, $a=5$, $b=2$ | M1, A1 | A1: correct equation in $x$ with no fractional terms e.g. $x\sqrt{16}+10\sqrt{2}=6x$. M1: attempt to solve linear equation producing answer of form $a\sqrt{2}$ or $a\sqrt{50}$. A1: $5\sqrt{2}$ oe (accept $1\sqrt{50}$) |

**Method 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\sqrt{2}x + 10 = 3\sqrt{2}x$ | M1, A1 | M1: writing $\sqrt{8}$ as $2\sqrt{2}$ or $\frac{6}{\sqrt{2}}$ as $3\sqrt{2}$. A1: correct equation with no fractional terms e.g. $2\sqrt{2}x+10=3\sqrt{2}x$ or $x\sqrt{8}+10=3\sqrt{2}x$ |
| $\sqrt{2}x = 10 \Rightarrow x = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ oe | M1, A1 | M1: attempt to solve producing answer of form $a\sqrt{2}$ or $a\sqrt{50}$. A1: $5\sqrt{2}$ oe, accept $1\sqrt{50}$ |

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5. Solve the equation

$$10 + x \sqrt { 8 } = \frac { 6 x } { \sqrt { 2 } }$$

Give your answer in the form $a \sqrt { } b$ where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{Edexcel C1 2014 Q5 [4]}}