Q.

The equation of smallest degree with real coefficients having 2+3i as one of the roots is

1 x2+4x+13=0  
2 x2+4x-13=0  
3 x2-4x+13=0  
4 x2-4x-13=0  

Ans.

(3)

Since complex roots always exist in pairs

So other root will be 2-3i

S=sum of roots=2+3i+2-3i=4

P=Product of roots=(2+3i)(2-3i)=4+9=13

So, required equation will be

           x2-Sx+P=0⇒x2-4x+13=0