Question

.If x² – 12x + 33 = 0, then what is the value of (x – 4)² + [1/(x – 4)²]​

Answer 1

First Factorise the given equation to get the value of X.

Solving by Shridharacharya Formula, we get x =( 6 + √3 ) or

( 6 - √3 ).

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Let ( x - 4 ) = K

${(x - 4)}^{2} + { (\frac{1}{x - 4}) }^{2} \\ \\ = {k }^{2} + { \frac{1}{k} }^{2} \\ \\ = ({k \: + \: \frac{1}{k}) }^{2} - 2$

K = 6 + √3 - 4 = 2 + √3

$({(2 + \sqrt{3}) \: + \: ( \frac{1}{2 \: + \: \sqrt{3} } )})^{2}$

$\frac{1}{2 \: + \sqrt{3} } \\ \\ = \frac{2 \: - \: \sqrt{3} }{(2 + \sqrt{3)}(2 - \sqrt{3}) } \\ \\ = 2 - \sqrt{3}$

Hence, The value in the first case is 14.

Similarly, in the second case, the value is 14.