Find-square-root-of-7-30-2-i-

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

$${Find}\:{square}\:{root}\:{of}\:\mathrm{7}−\mathrm{30}\sqrt{\mathrm{2}}{i}\:. \\ $$

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

Squareroot of 7−30(√2)i ? Let ±(√(7−30(√2) i))=p+q(√2) i where p,q∈Q (±(√(7−30(√2) i)))^2 =(p+q(√2) i)^2 7−30(√2) i=p^2 −2q^2 +2pq(√2) i p^2 −2q^2 =7 ∧ 2pq=−30 q=−((15)/p) p^2 −2(−((15)/p))^2 =7 p^2 −((450)/p^2 )=7 p^4 −7p^2 −450=0 p^2 =((7±(√(49+1800)))/2) =((7±43)/2)=25,−18 p=±5 or p=±3(√2) i ∉Q(discardable ) q=−((15)/(±5))=∓3 p+q(√2) i=±5∓3(√2) i =5−3(√2) i or −5+3(√2) i Squareroot of 7−30(√(2 )) i are 5−3(√2) i or −5+3(√2) i

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

Commented by Tinkutara last updated on 13/Jan/18

Thank you very much Sir! I got the answer.

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