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

[CaptainBlack]

(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]

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