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Question

If the quadratic equation z2+(2+4i)z+(c+5i)=0 where c is non zero real number, has real roots then the value of c equals

A
1615
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B
6516
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C
1516
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D
None of these
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Solution

The correct option is B 1516
z2+(2+4i)z+c+5i=0 ...(1)
Let a be the root of the equation (1).
Therefore, a2+(2+4i)a+c+5i=0 ...(2)
Since, a is real.
Therefore, a=¯a
Taking conjugate of equation (2) we get,
¯a2+(24i)¯a+c5i=0
a2+(24i)a+c5i=0 ...(3)
Substracting (2) & (3) we get,
8ia+10i=0a=54
Substituting a=54 in equation (2) we get,
(54)2(24i)54+c5i=0
c=1516
Ans: C

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