Question

If $n^2-16n-59=0$, then $(n-6{)}^2+\left(\frac{1}{(n-6{)}^2}\right)=?$

Answer 1

Given equation is $n^2-16n-59=0$

$n^2-12n-4n-36+36-59=0$

$n^2+36-12n-4n-95=0$

$(n-6{)}^2=4n+95$

we have to find out the value of $(n-6)^2+(\frac{1}{(n-6)^2})$

Now put the value of $(n-6{)}^2=4n+95$

$4n+95+\frac{1}{4n+95}$

$\frac{(4n+95{)}^2+1}{4n+95}$

$\frac{16n^2+1025+760n+1}{4n+95}$

$\frac{16n^2+1026+760n}{4n+95}$ Ans