Express (-sqrt{3}+sqrt{-2})(2 sqrt{3}-i) in t | vedclass

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Question

(A) Given expression: $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$Since $\sqrt{-2} = \sqrt{2}i$,the expression becomes $(-\sqrt{3}+\sqrt{2}i)(2\sqrt{3}-i)$.E

Vedclass · Accepted Answer

$(-6+\sqrt{2}) + i\sqrt{3}(1+2\sqrt{2})$

Vedclass · Answer

(A) Given expression: $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$Since $\sqrt{-2} = \sqrt{2}i$,the expression becomes $(-\sqrt{3}+\sqrt{2}i)(2\sqrt{3}-i)$.Expanding the product:$= (-\sqrt{3})(2\sqrt{3}) - (-\sqrt{3})(i) + (\sqrt{2}i)(2\sqrt{3}) - (\sqrt{2}i)(i)$$= -6 + \sqrt{3}i + 2\sqrt{6}i - \sqrt{2}i^2$Since $i^2 = -1$,we have:$= -6 + \sqrt{3}i + 2\sqrt{6}i + \sqrt{2}$$= (-6+\sqrt{2}) + i(\sqrt{3} + 2\sqrt{6})$$= (-6+\sqrt{2}) + i\sqrt{3}(1 + 2\sqrt{2})$

Express $(-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)$ in the form of $a+ib$.

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