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If $x^2-16x-59 = 0$, then what is the value of $(x-6)^2+[\frac{1}{(x-6)^2}]$?
Given: Equation: $x^2 – 16x – 59 = 0$.
To do: To find the value of $(x – 6)^2+[\frac{1}{(x – 6)^2}]$.
Solution:
Let $(x-6)=t$
$\Rightarrow x=t+6\ \ ........( 1)$
$x^2-16x+59=0$
Putting value of $x$ from equation $( 1)$
$\Rightarrow (t+6)^2-16(t+6)+59=0$
$\Rightarrow t^2+36+12t-16t-96+59=0$
$\Rightarrow t^2-4t-1=0$
On dividing the above equation by $t$,
$\Rightarrow t-4-\frac{1}{t}=0$
$\Rightarrow t-\frac{1}{t}=4$
On squaring above equation,
$\Rightarrow (t-\frac{1}{t})^2=4^2$
$\Rightarrow t^2+\frac{1}{t^2}-2=16$
$\Rightarrow t^2+\frac{1}{t}^2=18$
$\therefore (x-6)^2+[\frac{1}{(x-6)^2}]=18$.
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