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 1516z2+(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=¯aTaking conjugate of equation (2) we get,¯a2+(2−4i)¯a+c−5i=0⇒a2+(2−4i)a+c−5i=0 ...(3)Substracting (2) & (3) we get,8ia+10i=0⇒a=−54Substituting a=−54 in equation (2) we get, ⇒(−54)2−(2−4i)54+c−5i=0∴c=1516Ans: C

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