A) \[{{x}^{2}}=0\]
B) \[{{x}^{2}}+2bx+4=0\]
C) \[{{x}^{2}}-2bx+4=0\]
D) \[{{x}^{2}}-bx+1=0\]
Correct Answer: C
Solution :
Since, \[\alpha \] and\[\beta \] are roots of\[{{x}^{2}}+bx+1=0\] \[\therefore \] \[\alpha +\beta =-b,\alpha \beta =1\] Now, \[\left( -\alpha -\frac{1}{\beta } \right)+\left( -\beta -\frac{1}{\alpha } \right)\] \[=-(\alpha +\beta )-\left( \frac{1}{\beta }+\frac{1}{\alpha } \right)=-(\alpha +\beta )-\frac{(\alpha +\beta )}{\alpha \beta }\] \[=b+b=2b\] and \[\left( -\alpha -\frac{1}{\beta } \right)\left( -\beta -\frac{1}{\alpha } \right)\] \[=\alpha \beta +2+\frac{1}{\alpha \beta }=1+2+1=4\] Thus, the equation whose roots are \[-\alpha -\frac{1}{\beta }\] and\[-\beta -\frac{1}{\alpha },\] is \[{{x}^{2}}-2bx+4=0\].
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