Question

lf $\alpha ,\beta$ are the roots of $3x^2+5x-7=0$, then the value of ${\left(\frac{1}{3\alpha +5}\right)}^2+{\left(\frac{1}{3\beta +5}\right)}^2$ is

• A. $\frac{67}{63}$
• B. $\frac{67}{441}$
• C. $\frac{109}{441}$
• D. $\frac{109}{63}$

Answer 1

Answer: (B)

$\frac{67}{441}$

$Since\alpha ,\beta aretherootsoftheequation3x^2+5x-7\Rightarrow \alpha +\beta =-\frac{5}{3}and\alpha \beta =-\frac{7}{3}.Nowtheequationcanbewrittenas3{\alpha }^2+5\alpha -7=0\Rightarrow 3\alpha +5=\frac{7}{\alpha }----\left(1\right)and3{\beta }^2+5\beta -7=0\Rightarrow 3\beta +5=\frac{7}{\beta }\therefore Thegivenexpressionbecomes{\left(\frac{1}{\frac{7}{\alpha }}\right)}^2+{\left(\frac{1}{\frac{7}{\beta }}\right)}^2=\frac{{\alpha }^2+{\beta }^2}{49}=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{49}=\frac{{(-\frac{5}{3})}^2-2.(-\frac{7}{3})}{49}=\frac{67}{441}\therefore OptionBistheanswer.$