If $x^2 – 12 x + 33 = 0$, then what is the value of $(x – 4) ^2 + ( \frac{1}{(x – 4)})^2$.

Given: $x^2 – 12 x + 33 = 0$.

To do: To find the value of $(x – 4) ^2 + ( \frac{1}{(x – 4)})^2$.

Solution:

Let  $x=t+4$ 

$\Rightarrow (t+4)^2–12(t+4)+33=0$ 

$\Rightarrow t^2–8t+16–12t−48+33=0$ 

$\Rightarrow t^2–4t+1=0$ 

$\Rightarrow t+\frac{1}{t}=4$ 

Squaring both sides,

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

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

$\Rightarrow t^2+\frac{1}{t^2}=16-2=14$

On putting $t=x-4$, it becomes

$(x−4)^2+\frac{1}{(x−4)^2}=14$ 
 

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Updated on: 10-Oct-2022

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