Question

$\alpha ,\beta$ are roots of $x^2+6x+4=0$ then value of ${(\alpha -6)}^{-2}+{(\beta -6)}^{-2}$ is ?

• A. $\frac{7}{4}$
• B. $2$
• C. $4$
• D. $\frac{43}{1444}$

Answer 1

The correct option is D $\frac{43}{1444}$

$x^2+6x+4(\alpha ,\beta )$ roots

$x^2+6x+4=(x-\alpha )(x-\beta )$

$6^2+6.6+4=(\alpha -6)(\beta -6)$

$76=(\alpha -6)(\beta -6)$

$(\alpha -6{)}^{-2}+(\beta -6{)}^{-2}$

$\frac{1}{(\alpha -6{)}^{-2}}+\frac{1}{(\beta -6{)}^2}$

$\frac{(\alpha -6{)}^2+(\beta -6{)}^2}{{\left[\right(\alpha -6\left)\right(\beta -6\left)\right]}^2}$

$\frac{{\alpha }^2+{\beta }^2+72-12(\alpha +\beta )}{(76{)}^2}$

$\frac{({\alpha }^2+{\beta }^2+2\alpha \beta )+72-12(\alpha +\beta )-2\alpha \beta }{(76{)}^2}$

$\frac{(-6{)}^2+72-12(-6)-2(4)}{(76{)}^2}$

$=\frac{43}{38.38}$