A) 10
B) 22
C) 30
D) 29
E) 31
Correct Answer: E
Solution :
Since,\[\frac{3}{2}+\frac{7}{2}i\]is a solution of equation \[a{{x}^{2}}-6x+b=0\] \[\therefore \] \[a{{\left( \frac{3}{2}+\frac{7}{2}i \right)}^{2}}-\left( \frac{3}{2}+\frac{7}{2} \right)i+b=0\] \[\Rightarrow \] \[a\left( -10+\frac{21}{2}i \right)-6\left( \frac{3}{2}+\frac{7}{2}i \right)+b=0\] \[\Rightarrow \] \[10a-b=-9\]and \[\frac{21}{2}a-21=0\] \[\Rightarrow \] \[a=2\] \[\therefore \] \[b=29\] \[\therefore \] \[a+b=2+29=31\] Alternate Let\[\alpha \]and\[\beta \]are the roots of given equation. Let \[\alpha =\frac{3}{2}+\frac{7}{2}i\] Then, \[\beta =\frac{3}{2}-\frac{7}{2}i\] and given equation is \[a{{x}^{2}}-6x+b\] Now, \[\alpha +\beta \,=\frac{6}{a}=\frac{6}{2}\Rightarrow a=2\] \[\alpha .\beta =\frac{b}{a}=\left( \frac{3}{2}+\frac{7}{2}i \right)\left( \frac{3}{2}-\frac{7}{2}i \right)=\frac{9}{4}+\frac{49}{4}\] \[\Rightarrow \] \[b=29\] \[\therefore \] \[a+b=29+2=31\]
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