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Exercise 1.2 · Q73

Q.Solve the following quadratic equation : x2−(2+i)x−(1−7i)=0x^2-(2+i)x-(1-7i)=0

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For x2−(2+i)x−(1−7i)=0x^2-(2+i)x-(1-7i)=0: a=1, b=−(2+i), c=−(1−7i)a=1,\ b=-(2+i),\ c=-(1-7i). D=b2−4ac=(2+i)2−4(1)(−(1−7i))=(4+4i+i2)+4(1−7i)=(3+4i)+(4−28i)=7−24iD=b^2-4ac=(2+i)^2-4(1)(-(1-7i))=(4+4i+i^2)+4(1-7i)=(3+4i)+(4-28i)=7-24i. Find sqrt7−24i\\sqrt{7-24i}: set =p+iq=p+iq, so p2−q2=7,2pq=−24p^2-q^2=7,\\ 2pq=-24. (p2+q2)2=49+576=625Rightarrowp2+q2=25(p^2+q^2)^2=49+576=625\\Rightarrow p^2+q^2=25. Then 2p2=32Rightarrowp=pm42p^2=32\\Rightarrow p=\\pm4, 2q2=18Rightarrowq=pm32q^2=18\\Rightarrow q=\\pm3, opposite signs (since 2pq<02pq<0): (4,−3)(4,-3) or (−4,3)(-4,3), so $\sqrt{7-24i}= …

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