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Question

Let P(z)=z3+az2+bz+c, where a,b,cR. If there exists a complex number w such that the three roots of P(z) are w+3i,w+9i and 2w4, where i2=1, then the value of a+b+c is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

A
136
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B
136
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C
280
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D
280
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Solution

The correct option is B 136
Let w=x+iy, so the three roots are
x1=x+i(y+3)x2=x+i(y+9)x3=(2x4)+2iy
Now,
P(z)=z3+az2+bz+c
Sum of roots
x1+x2+x3=a
As aR, so
Im(x1+x2+x3)=04y+12=0y=3
Now,
x1=xx2=x+6ix3=(2x4)6i
Product of roots
x1x2x3=c
As cR, so
Im(x1x2x3)=0Im(x(x+6i)((2x4)6i))=0(2x4)×66x=0x=4
Now,
x1=4x2=4+6ix3=46i
Therefore,
a=12,c=208,b=84
Hence, a+b+c=136

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