Question : If $x^{2}-12x+33=0$, then what is the value of $(x-4)^{2}+\frac{1}{(x-4)^{2}}?$
Option 1: $16$
Option 2: $14$
Option 3: $18$
Option 4: $20$
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $14$
Solution : Put $(x-4) = m$ ⇒ $x = m+4$ ⇒ $(m + 4)^{2} - 12(m + 4) + 33 = 0$ ⇒ $m^{2} + 16 + 8m - 12m - 48 + 33 = 0$ ⇒ $m^{2}- 4m + 1 = 0$ On dividing the equation by $m$, ⇒ $m + \frac{1}{\text{m}} = 4$ Now putting $(x - 4) = m$ in $(x - 4)^{2} + [\frac{1}{(x - 4)^{2}}]$ ⇒ $m^{2} + \frac{1}{m^{2}} = (m +\frac{1}{m})^{2}-2$ ⇒ $m^{2} + \frac{1}{m^{2}} = 16 -2 = 14$ So, $(x-4)^{2}+\frac{1}{(x-4)^{2}}=14$ Hence, the correct answer is $14$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile