Question 1 Report
(a) Given that \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = a + b\sqrt{6}\), find a and b.
(b) If \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\), find y.
(a) Expand \((\sqrt{3}-5\sqrt{2})(\sqrt{3}+\sqrt{2})\):
\(=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{3}-5\sqrt{2}\cdot\sqrt{2}\)
\(=3+\sqrt{6}-5\sqrt{6}-10=-7-4\sqrt{6}\).
Comparing with \(a+b\sqrt{6}\): \(\mathbf{a=-7,\ b=-4}\).
(b) \(\dfrac{2^{1-y}\times 2^{y-1}}{2^{y+2}}=8^{2-3y}\).
Numerator: \(2^{(1-y)+(y-1)}=2^{0}=1\). Left side \(=2^{-(y+2)}\).
Right side: \(8^{2-3y}=2^{3(2-3y)}=2^{6-9y}\).
\(-(y+2)=6-9y \Rightarrow -y-2=6-9y \Rightarrow 8y=8 \Rightarrow y=\mathbf{1}\).
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