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Exercise 1.2 · Q74

Q.Solve the following quadratic equation : x2−(32+2i)x+62 i=0x^2-(3\sqrt2+2i)x+6\sqrt2\,i=0

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For x2−(3sqrt2+2i)x+6sqrt2,i=0x^2-(3\\sqrt2+2i)x+6\\sqrt2\\,i=0: a=1, b=−(3sqrt2+2i), c=6sqrt2,ia=1,\ b=-(3\\sqrt2+2i),\ c=6\\sqrt2\\,i. D=(3sqrt2+2i)2−4(6sqrt2,i)=(18+12sqrt2,i+4i2)−24sqrt2,i=(18−4+12sqrt2i)−24sqrt2i=14−12sqrt2iD=(3\\sqrt2+2i)^2-4(6\\sqrt2\\,i)=(18+12\\sqrt2\\,i+4i^2)-24\\sqrt2\\,i=(18-4+12\\sqrt2i)-24\\sqrt2i=14-12\\sqrt2i. Rather than take a square root from scratch, notice this factors: since the two roots multiply to c=6sqrt2,ic=6\\sqrt2\\,i and add to b′=3sqrt2+2ib'=3\\sqrt2+2i, try x=3sqrt2x=3\\sqrt2 and x=2ix=2i directly — …

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