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Question Number 208892 by hardmath last updated on 26/Jun/24

$$\mathrm{Find}:$$$$\sqrt{-16}\cdot \sqrt{-9}=?$$

Commented by Adeyemi889 last updated on 26/Jun/24

$$$$$$\sqrt{-16}=\sqrt{\left(16\right)(-1)}=\sqrt{-1}\times \sqrt{16}$$$$note\sqrt{-1}=\mathit{i}$$$$\sqrt{-16}=4\mathit{i}$$$$$$

Answered by mr W last updated on 26/Jun/24

$$=\sqrt{-1}\times \sqrt{16}\times \sqrt{-1}\times \sqrt{9}$$$$={\left(\sqrt{-1}\right)}^2\times 4\times 3$$$$=-12$$

Answered by Frix last updated on 26/Jun/24

$$\sqrt{-16}=\sqrt{{16\mathrm{e}}^{\mathrm{i}\pi }}={4\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$$\sqrt{-9}=\sqrt{{9\mathrm{e}}^{\mathrm{i}\pi }}={3\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$$\sqrt{-16}\times \sqrt{-9}={4\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}\times {3\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}={12\mathrm{e}}^{\mathrm{i}\pi }=-12$$

Commented by hardmath last updated on 28/Jun/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professors}$$

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