If x2 - 12x + 33 = 0 find the value of (x-4)2+\(\frac{1}{({x-4})^{2}}\) - 5

16 14 9 20

9

Let x - 4 = t  x = t + 4 Put in given equation (t + 4)2 - 12 (t + 4) + 33 = 0 t2 + 16 + 8t - 12t - 48 + 33 = 0 t2 - 4t + 1 = 0 t+\(\frac{1}{t}\) = 4 t2+\(\frac{1}{{t}^{2}}\) - 5 = (4)2 - 2 - 5 = 9