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Question Number 27681 by ajfour last updated on 12/Jan/18

$$Findsquarerootof7-30\sqrt{2}i.$$

Commented by Rasheed.Sindhi last updated on 13/Jan/18

$$\mathrm{Squareroot}\mathrm{of}7-30\sqrt{2}i?$$$$\mathrm{Let}\pm \sqrt{7-30\sqrt{2}\mathrm{i}}=\mathrm{p}+\mathrm{q}\sqrt{2}\mathrm{i}$$$$\mathrm{where}\mathrm{p},\mathrm{q}\in \mathbb{Q}$$$${(\pm \sqrt{7-30\sqrt{2}\mathrm{i}})}^2={(\mathrm{p}+\mathrm{q}\sqrt{2}\mathrm{i})}^2$$$$7-30\sqrt{2}\mathrm{i}={\mathrm{p}}^2-{2\mathrm{q}}^2+2\mathrm{pq}\sqrt{2}\mathrm{i}$$$${\mathrm{p}}^2-{2\mathrm{q}}^2=7\wedge 2\mathrm{pq}=-30$$$$\mathrm{q}=-\frac{15}{\mathrm{p}}$$$${\mathrm{p}}^2-2{(-\frac{15}{\mathrm{p}})}^2=7$$$${\mathrm{p}}^2-\frac{450}{{\mathrm{p}}^2}=7$$$${\mathrm{p}}^4-{7\mathrm{p}}^2-450=0$$$${\mathrm{p}}^2=\frac{7\pm \sqrt{49+1800}}{2}$$$$=\frac{7\pm 43}{2}=25,-18$$$$\mathrm{p}=\pm 5\mathrm{or}\mathrm{p}=\pm 3\sqrt{2}\mathrm{i}\notin \mathbb{Q}\left(\mathrm{discardable}\right)$$$$\mathrm{q}=-\frac{15}{\pm 5}=\mp 3$$$$\mathrm{p}+\mathrm{q}\sqrt{2}\mathrm{i}=\pm 5\mp 3\sqrt{2}\mathrm{i}$$$$=5-3\sqrt{2}\mathrm{i}\mathrm{or}-5+3\sqrt{2}\mathrm{i}$$$$\mathrm{Squareroot}\mathrm{of}7-30\sqrt{2}\mathrm{i}\mathrm{are}$$$$5-3\sqrt{2}\mathrm{i}\mathrm{or}-5+3\sqrt{2}\mathrm{i}$$

Commented by Tinkutara last updated on 13/Jan/18

Thank you very much Sir! I got the answer.

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