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

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $\alpha + \beta = 7$ and $\alpha\beta = 1$
$	herefore\ \alpha - 7 = - \beta,\ \ \ \ \ \beta - 7 = - \alpha$
$	herefore\ \ \ \ \ \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}}$
$= (\alpha + \beta)^{2} - 2\alpha\beta$
$= 49 - 2 = 47$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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