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,

$\Rightarrow (t-\frac{1}{t})^2=4^2$

$\Rightarrow t^2+\frac{1}{t^2}-2=16$

$\Rightarrow t^2+\frac{1}{t}^2=18$

$\therefore (x-6)^2+[\frac{1}{(x-6)^2}]=18$.
 

Updated on: 10-Oct-2022

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