Question

If $\alpha$ and $\beta$ are roots of the equation $x^2-15x+1=0$, then the value of ${(\frac{1}{\alpha }-15)}^{-2}+{(\frac{1}{\beta }-15)}^{-2}$ is

• A. $225$
• B. $900$
• C. $223$
• D. None of these

Answer 1

Answer: (C)

$223$

$\alpha$ and $\beta$ are the roots of $x^2-15x+1=0$

$\therefore \alpha +\beta =15,\alpha \beta =1$

Now ${(\frac{1}{\alpha }-15)}^{-2}+{(\frac{1}{\beta }-15)}^{-2}={(-\alpha )}^{-2}+{(-\beta )}^{-2}$ $(\because \frac{1}{\alpha }=\beta ,\frac{1}{\beta }=\alpha ,\alpha +\beta =15)$

$={\left(\frac{1}{-\alpha }\right)}^2+{\left(\frac{1}{-\beta }\right)}^2=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}={\alpha }^2+{\beta }^2$ $(\because \alpha \beta =1)$

$={(\alpha +\beta )}^2-2\alpha \beta$

$=225-2=223$