A) \[16.25\]
B) \[17.25\]
C) 18
D) 17
Correct Answer: B
Solution :
\[2x+\frac{1}{3x}=3\] Multiply both side by \[\frac{3}{2},\] \[2x\times \frac{3}{2}+\frac{3}{2}\times \frac{1}{3x}=3\times \frac{3}{2}\,\,3x+\frac{1}{2x}=\frac{9}{2}\] Taking square of both side- \[{{\left( 3x+\frac{1}{2x} \right)}^{2}}={{\left( \frac{9}{2} \right)}^{2}}\Rightarrow 9{{x}^{2}}+\frac{1}{4{{x}^{2}}}+2\times 3x\times \frac{1}{2x}\] \[=\frac{81}{4}\Rightarrow 9{{x}^{2}}+\frac{1}{4{{x}^{2}}}=\frac{81}{4}-3\Rightarrow \frac{81-12}{4}\] \[\underline{\mathbf{=17}\mathbf{.25}}\]
You need to login to perform this action.
You will be redirected in
3 sec
