A) \[{{\alpha }^{{}^{3}/{}_{2}}}\]and\[{{\beta }^{{}^{3}/{}_{2}}}\]
B) \[\alpha {{\beta }^{{}^{1}/{}_{2}}}\]and\[{{\alpha }^{{}^{1}/{}_{2}}}\beta \]
C) \[\sqrt{\alpha \beta }\]and\[\alpha \beta \]
D) \[{{\alpha }^{-\frac{3}{2}}}\]and\[{{\beta }^{-\frac{3}{2}}}\]
Correct Answer: A
Solution :
Let\[\frac{1}{\sqrt{\alpha }}\]and\[\frac{1}{\sqrt{\beta }}\]be the roots of \[a{{x}^{2}}+bx+1=0\] \[\frac{1}{\sqrt{\alpha }}+\frac{1}{\sqrt{\beta }}=\left( \frac{\sqrt{\alpha }+\sqrt{\beta }}{\sqrt{\alpha \beta }} \right)=-\frac{b}{a}\] \[\frac{1}{\sqrt{\alpha }\sqrt{\beta }}=\frac{1}{a}\Rightarrow a=\sqrt{\alpha \beta }\] \[b=-\left( \sqrt{\alpha }+\sqrt{\beta } \right)\] \[x\left( x+{{b}^{3}} \right)+({{a}^{3}}-3abx)=0\] \[\Rightarrow \]\[{{x}^{2}}+({{b}^{3}}-3ab)x+{{a}^{3}}=0\] Putting values of a and b, we get \[{{x}^{2}}+\left[ {{\left( -\sqrt{\alpha }+\sqrt{\beta } \right)}^{3}}+3\left( \sqrt{\alpha \beta } \right)\left( \sqrt{\alpha }+\beta \right) \right]+{{(\alpha \beta )}^{3/2}}=0\]\[\Rightarrow \]\[{{x}^{2}}-\left[ {{\alpha }^{3/2}}+{{\beta }^{3/2}}+3\sqrt{\alpha \beta }\left( \sqrt{\alpha }+\sqrt{\beta } \right)-3\sqrt{\alpha \beta }(\sqrt{\alpha }+\sqrt{\beta }) \right]\]\[x+{{(\alpha \beta )}^{3/2}}=0\] \[\Rightarrow \]\[{{x}^{2}}-({{\alpha }^{3/2}}+{{\beta }^{3/2}})x+{{\alpha }^{3/2}}{{\beta }^{3/2}}=0\] Roots of this equation are \[{{\alpha }^{3/2}}+,{{\beta }^{3/2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
