Question

If $x+\dfrac{1}{x-1}=5$, then find the value of $\,(x-1)^2+\dfrac{1}{(x-1)^2}\,$?

• A. $10$
• B. $11$
• C. $14$
• D. None of these

Hint:

When you see $x$ together with $\frac{1}{x-1}$ and are asked for $(x-1)^2+\frac{1}{(x-1)^2}$, set $y=x-1$ and use $(y+\frac{1}{y})^2=y^2+\frac{1}{y^2}+2$.

Answer 1

The Correct Option is C

Step 1: Substitute $y=x-1$.
Then $x=y+1$, so \[ x+\frac{1}{x-1}= (y+1)+\frac{1}{y}=5 \ \Rightarrow\ y+\frac{1}{y}=4. \] Step 2: Use the identity for squares.
\[ \left(y+\frac{1}{y}\right)^2 = y^2+\frac{1}{y^2}+2 \Rightarrow y^2+\frac{1}{y^2} = 4^2-2=16-2=14. \] Step 3: Translate back.
Since $y=x-1$, we get \[ (x-1)^2+\frac{1}{(x-1)^2}=14. \] \[ \boxed{14} \]