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Question

Solve x2(7i)x+(18i)=0 .

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Solution

x2(7i)x+(18i)=0
equation is in the form ax2+bx+c=0

a=1,b=(7i),c=18i

α=b+b24ac2a and β=bb22ac2a

α=(7i)+(7i)24(18i)2 β=(7i)(7i)24(18i)2

α=(7i)2410i2 β=(7i)2410i2

Now for the value of 2410i

Let a+ib=2410i

a2b2+2abi=2410i

a2b2=242ab=10

Consider

(a2+b2)2=(a2b2)2+4a2b2

=(24)2+(10)2=6+6

a2+b2=26

Now,

a2b2=24 and a2+b2=26

a2=1b2=25

a=±1b=±5

2ab=10 So, a and b must be opposite signs

a=1 and b=5 or a=1 and b=5

24100=15i or 1+5i

Let us consider 15i

α=7i+(15i)2 and β=(7i)(15i)2

α=43iβ=33i

Roots are 43i and 33i

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