A) \[-1\]
B) 1
C) 0
D) does not exist
Correct Answer: B
Solution :
\[{{z}_{1}}+{{z}_{2}}=-1,\,\,{{z}_{1}}{{z}_{2}}=\frac{b}{3}\] \[{{0}^{2}}+{{z}_{1}}^{2}+{{z}_{2}}^{2}=0\times {{z}_{1}}+0\times {{z}_{2}}+{{z}_{1}}{{z}_{2}}\] \[\Rightarrow \,\,\,{{({{z}_{1}}+{{z}_{2}})}^{2}}-2{{z}_{1}}{{z}_{2}}={{z}_{1}}{{z}_{2}}\Rightarrow 1=3{{z}_{1}}{{z}_{2}}=3\frac{b}{3}\] \[\Rightarrow \,\,\,b=1\]You need to login to perform this action.
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