Easy -1.3 This is a straightforward rationalizing the denominator question requiring only standard techniques: recognizing difference of two squares in part (a), then multiplying numerator and denominator by the conjugate in part (b). Both parts are routine textbook exercises with no problem-solving or insight required, making it easier than average.
\(16+4\sqrt{3}-4\sqrt{3}-(\sqrt{3})^2\) or \(16-3 = 13\)
M1, A1c.a.o
For 4 terms, at least 3 correct. e.g. \(8+4\sqrt{3}-4\sqrt{3}-(\sqrt{3})^2\) or \(16\pm8\sqrt{3}-(\sqrt{3})^2\) or \(16+3\). \(4^2\) instead of 16 is OK. \((4+\sqrt{3})(4+\sqrt{3})\) scores M0A0
For a correct attempt to rationalise the denominator. Can be implied. NB \(\frac{-4+\sqrt{3}}{-4+\sqrt{3}}\) is OK
\(=\frac{26(4-\sqrt{3})}{13}=8-2\sqrt{3}\) or \(8+(-2)\sqrt{3}\) or \(a=8\) and \(b=-2\)
A1
(2)
## Question 6:
### Part (a):
| $16+4\sqrt{3}-4\sqrt{3}-(\sqrt{3})^2$ or $16-3 = 13$ | M1, A1c.a.o | For 4 terms, at least 3 correct. e.g. $8+4\sqrt{3}-4\sqrt{3}-(\sqrt{3})^2$ or $16\pm8\sqrt{3}-(\sqrt{3})^2$ or $16+3$. $4^2$ instead of 16 is OK. $(4+\sqrt{3})(4+\sqrt{3})$ scores M0A0 |
### Part (b):
| $\frac{26}{4+\sqrt{3}}\times\frac{4-\sqrt{3}}{4-\sqrt{3}}$ | M1 | For a correct attempt to rationalise the denominator. Can be implied. NB $\frac{-4+\sqrt{3}}{-4+\sqrt{3}}$ is OK |
| $=\frac{26(4-\sqrt{3})}{13}=8-2\sqrt{3}$ or $8+(-2)\sqrt{3}$ or $a=8$ and $b=-2$ | A1 | (2) |
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