Question

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

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

Answer 1

The correct option is B

$47$

Explanation for the correct option:

Find the value of $\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}{\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{-}\mathbf{}\mathbf{7}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}{\mathbf{(}\mathbf{\beta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{7}\mathbf{)}}^{\mathbf{2}}$ :

Given that $\alpha$ and $\beta$ are the roots of the equation $x^2-7x+1=0$.

$\Rightarrow (\alpha +\beta )=7;\alpha .\beta =1$

and $\alpha –7=–\beta ;\beta –7=–\alpha$

Now, $1/{(\mathrm{\alpha }-7)}^2+1/{(\mathrm{\beta }-7)}^2$

$$\begin{gathered} =\frac{1}{{(-\beta )}^2}+\frac{1}{{(-\alpha )}^2} \\ =\frac{1}{{\beta }^2}+\frac{1}{{\alpha }^2} \\ =\frac{{\alpha }^2+{\beta }^2}{{(\alpha .\beta )}^2} \\ =\frac{{\alpha }^2+{\beta }^2+2\alpha .\beta -2\alpha .\beta }{{(\alpha .\beta )}^2} \\ =\frac{{(\alpha +\beta )}^2-2\alpha .\beta }{{(\alpha .\beta )}^2} \\ =\frac{49-2}{1} \\ =47 \end{gathered}$$

Hence, the correct option is (B).