A) \[\sqrt{2}\]
B) 2
C) \[\sqrt{3}\]
D) 3
Correct Answer: C
Solution :
\[{{x}^{2}}+bx-1=0\And {{x}^{2}}+x+b=0\]have common root \[\alpha .\] \[\Rightarrow \]\[{{\alpha }^{2}}+b\alpha -1=0\] \[{{\alpha }^{2}}+\alpha +b=0\] \[\Rightarrow \]\[\frac{{{\alpha }^{2}}}{{{b}^{2}}+1}=\frac{\alpha }{-(b+1)}=\frac{1}{(1-b)}\] \[\Rightarrow \]\[{{(b+1)}^{2}}=({{b}^{2}}+1)(1-b)\] \[\Rightarrow \]\[{{b}^{2}}+2b+1={{b}^{2}}-{{b}^{3}}+1-b\] \[\Rightarrow \]\[{{b}^{3}}+3b=0\] \[\Rightarrow \]\[b=0\]or\[{{b}^{2}}=-3\] \[\Rightarrow \]\[\Rightarrow \]\[\Rightarrow \] when b = 0 then common roots is (.1) hence b = 0 rejected. so \[{{b}^{2}}=-3\]\[\Rightarrow \]\[b=\pm \sqrt{3}i\]\[\Rightarrow \]\[|b|=\sqrt{3}\]
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