Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-4x+1=0$, then the value of $\frac{1}{\alpha +2\beta }+\frac{1}{\beta +2\alpha }$ is _________.

• A. $\frac{12}{13}$
• B. $\frac{6}{5}$
• C. $\frac{8}{15}$
• D. $\frac{12}{17}$

Answer 1

The correct option is D $\frac{12}{17}$

$(\frac{1}{(\alpha +2\beta )}+\frac{1}{(\beta +2\alpha )})=\frac{\beta +2\alpha +2\beta +\alpha }{(\alpha +2\beta )(\beta +2\alpha )}$

$=\frac{3(\alpha +\beta )}{\alpha \beta +2{\beta }^2+2{\alpha }^2+4\alpha \beta }=\frac{3(\alpha +\beta )}{5\alpha \beta +2\left({\alpha }^2{+\beta }^2\right)}$

$=\frac{3(\alpha +\beta )}{5\alpha \beta +2({(\alpha +\beta )}^2-2\alpha \beta )}=\frac{3(\alpha +\beta )}{2{(\alpha +\beta )}^2+\alpha \beta }$ $\to \left(I\right)$

Since $\alpha ,\beta$ are the roots of $2x^2-4x+1=0$,

Sum of the roots $=\alpha +\beta =-\frac{b}{a}=-(-4)/2=2$ $\to \left(II\right)$

Product of the roots $=\alpha \beta =\frac{c}{a}=1/2$ $\to \left(III\right)$

Substituting $\left(II\right),\left(III\right)$ in $\left(I\right)$,

$(\frac{1}{(\alpha +2\beta )}+\frac{1}{(\beta +2\alpha )})=\frac{3\times 2}{2(2{)}^2+\frac{1}{2}}=\frac{6}{8-\frac{1}{2}}=\frac{12}{16+1}=\frac{12}{17}$

Hence, option D is correct.