# Question 2, AQA AS Maths Pure Core 1, May 2011

#### CaptainBlack

##### Well-known member
(a) (i) Express $$\sqrt{48}$$ in the form $$k\sqrt{3}$$, where $$k$$ is an integer. (1 mark)

.....(ii) Simplify $$\displaystyle \frac{\sqrt{48}+2\sqrt{27}}{\sqrt{12}}$$ giving your answer as an integer. (3 marks)

(b) Express $$\displaystyle \frac{1-5\sqrt{5}}{3+\sqrt{5}}$$ in the form $$m+n\sqrt{5}$$, where m and n are integers. (4 marks)

CB

#### CaptainBlack

##### Well-known member
Answer

(a) (i)
Since $$48=16 \times 3$$ we have: $$\sqrt{48}=\sqrt{16 \times 3}=4\sqrt{3}$$.

(a) (ii) $\frac{\sqrt{48}+2\sqrt{27}}{\sqrt{12}}=\frac{4 \sqrt{3}+2 \times 3 \sqrt{3}}{2 \sqrt{3}}=\frac{4+6}{2}=5$

(b) We multiply top and bottom by $$3-\sqrt{5}$$:

$\frac{1-5\sqrt{5}}{3+\sqrt{5}}=\frac{(1-5\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{3-15 \sqrt{5}- \sqrt{5}+5( \sqrt{5})^2}{9-5}=\frac{28-16\sqrt{5}}{4}=7-4\sqrt{5}$

CB