Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-7x+1=0,$ then the value of $\frac{1}{{(\alpha -7)}^2}+\frac{1}{{(\beta -7)}^2}$ is:

• A. 45
• B. 47
• C. 49
• D. 50
• E. 51

Answer 1

Answer: (B)

Given $\alpha$ and $\beta$ are the roots of $x^2-7x+1=0$ $\Rightarrow$ $\alpha +\beta =7$ and $\alpha \beta =1$ $\therefore$ $\alpha -7=\beta ,\beta -7=-\alpha$ $\therefore$ $\frac{1}{{(\alpha -7)}^2}+\frac{1}{{(\beta -7)}^2}=\frac{1}{{\beta }^2}+\frac{1}{{\alpha }^2}$ $=\frac{{\alpha }^2+{\beta }^2}{(\alpha {\beta }^2)}$ $={(\alpha +\beta )}^2-2\alpha \beta$ $=49-2=47$