x2−(3−2i)x−(2i−2)=0x=2(1)(3−2i)±(3−2i)2−4(1)(−(2i−2))==2(3−2i)±9−4−12i+8i−8==23−2i±−3−4i=23−2i±(1)2+(2i)2−2(1)(2i)=23−2i±(1−2i)⇒23−2i+1−2i or 23−2i−1+2i⇒2−2i or 1+0i So αγ+βδ=2(1)+(−2)(0)=2
If α+iβ and γ+iδ are the roots of x2−(3−2i)x−(2i−2)=0,i=−1, then αγ+βδ is equal to :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
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