(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) \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = \sqrt{3 \times 3} + \sqrt{3 \times 2} - 5\sqrt{2 \times 3} - 5\sqrt{2 \times 2}\)
= \(3 + \sqrt{6} - 5\sqrt{6} - 5(2)\)
= \(-7 - 4\sqrt{6}\)
\(\therefore\) a = -7 and b = -4.
(b) \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\)
\(\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\)
\(2^{0 - (y + 2)} = 2^{6 - 9y} \implies 2^{- y - 2} = 2^{6 - 9y}\)
\(- y - 2 = 6 - 9y \implies - y + 9y = 6 + 2\)
\(8y = 8 \implies y = 1\).
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