2 Express \(\frac { a + \sqrt { 2 } } { 3 - \sqrt { 2 } }\) in the form \(\mathrm { p } + \mathrm { q } \sqrt { 2 }\), giving \(p\) and \(q\) in terms of \(a\).
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Question 2:
Answer Marks
Guidance
Answer Marks
Guidance
\(\frac{(a+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\) M1
Attempt to rationalize the denominator
\(= \frac{3a + a\sqrt{2} + 3\sqrt{2} + 2}{9-2}\) M1
Attempt to expand the brackets
\(= \frac{3a+2}{7} + \frac{a+3}{7}\sqrt{2}\) A1 [3]
Must be in the form \(p + q\sqrt{2}\). Mark final answer. Allow for instead stating \(p = \frac{3a+2}{7}\), \(q = \frac{a+3}{7}\)
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## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(a+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$ | M1 | Attempt to rationalize the denominator |
| $= \frac{3a + a\sqrt{2} + 3\sqrt{2} + 2}{9-2}$ | M1 | Attempt to expand the brackets |
| $= \frac{3a+2}{7} + \frac{a+3}{7}\sqrt{2}$ | A1 [3] | Must be in the form $p + q\sqrt{2}$. Mark final answer. Allow for instead stating $p = \frac{3a+2}{7}$, $q = \frac{a+3}{7}$ |
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2 Express $\frac { a + \sqrt { 2 } } { 3 - \sqrt { 2 } }$ in the form $\mathrm { p } + \mathrm { q } \sqrt { 2 }$, giving $p$ and $q$ in terms of $a$.
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q2 [3]}}