Question

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

Answer 1

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

Here, $a=1,b=-7,c=1$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{1}{1}=1$ ----- ( 1 )

$\Rightarrow$ ${\alpha }^2{\beta }^2=1$ ----- ( 2 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=-\frac{-7}{1}=7$ ----- ( 3 )

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

$\Rightarrow$ $(7{)}^2={\alpha }^2+{\beta }^2+2(1)$ [ Using ( 1 ) and ( 3 ) ]

$\Rightarrow$ $49={\alpha }^2+{\beta }^2+2$

$\therefore$ ${\alpha }^2+{\beta }^2=47$ ------- ( 4 )

Now,

$\Rightarrow$ $\frac{1}{(\alpha -7{)}^2}+\frac{1}{(\beta -7{)}^2}=\frac{(\beta -7{)}^2+(\alpha -7{)}^2}{(\alpha -7{)}^2(\beta -7{)}^2}$

$=\frac{{\beta }^2-14\beta +49+{\alpha }^2-14\alpha +49}{({\beta }^2-14\beta +49)({\alpha }^2-14\alpha +49)}$

$=\frac{{\alpha }^2+{\beta }^2-14(\alpha +\beta )+98}{{\alpha }^2{\beta }^2-14\alpha {\beta }^2+49{\beta }^2-14{\alpha }^2\beta +196\alpha \beta -686\beta +49{\alpha }^2-686\alpha +2401}$

$=\frac{47-14\left(7\right)+98}{{\alpha }^2{\beta }^2-14\alpha \beta (\alpha +\beta )+49({\alpha }^2+{\beta }^2)+196\alpha \beta -686(\alpha +\beta )+2401}$ [ Using ( 3 ) and ( 4 ) ]

$=\frac{47}{1-14\left(1\right)\left(7\right)+49\left(47\right)+196\left(1\right)-686\left(7\right)+2401}$

$=\frac{47}{1-98+2303+196-4802+2401}$

$=\frac{47}{1}$

$\therefore$ $\frac{1}{(\alpha -7{)}^2}+\frac{1}{(\beta -7{)}^2}=47$