A) 0
B) 1
C) 2
D) \[2\sqrt{\frac{b}{a}}\]
Correct Answer: D
Solution :
\[{{\left( \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }} \right)}^{2}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }+2=\frac{{{(\alpha +\beta )}^{2}}}{\alpha \beta }\] \[=\frac{{{\left( -\frac{b}{a} \right)}^{2}}}{\left( \frac{b}{a} \right)}=\frac{b}{a}\] \[\therefore \,\,\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\sqrt{\frac{b}{a}}\] [\[\because \,\,\alpha ,\beta \] are real] \[\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}+\sqrt{\frac{b}{a}}=2\sqrt{\frac{b}{a}}\]
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