If x2 - 12x + 33 = 0, then what is the value of (x - 4)2 + $$\frac{1}{{{{\left( {x - 4} \right)}^2}}}?$$
A. 16
B. 14
C. 18
D. 20
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {x^2} - 12x + 33 = 0 \cr & {x^2} - 12x + 36 - 3 = 0 \cr & {\left( {x - 6} \right)^2} - 3 = 0 \cr & x = \sqrt 3 + 6 \cr & {\left( {x - 4} \right)^2} + \frac{1}{{{{\left( {x - 4} \right)}^2}}} \cr & \therefore {\left( {\sqrt 3 + 2} \right)^2} + \frac{1}{{{{\left( {\sqrt 3 + 2} \right)}^2}}} \cr & = {\left( {\sqrt 3 + 2} \right)^2} + {\left( {\sqrt 3 - 2} \right)^2} \cr & = 2\left( {{{\sqrt 3 }^2} + {2^2}} \right) \cr & = 2 \times 7 \cr & = 14 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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