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: Let \(y=x-1\). Then \(x+\dfrac{1}{x-1}=(y+1)+\dfrac{1}{y}=5\), so \(y+\dfrac{1}{y}=4\).

Step 2: Squaring both sides, \(\left(y+\dfrac{1}{y}\right)^2=y^2+\dfrac{1}{y^2}+2\), so \(16=y^2+\dfrac{1}{y^2}+2\).

Step 3: This gives \(y^2+\dfrac{1}{y^2}=16-2=14\), and since \(y=x-1\), \((x-1)^2+\dfrac{1}{(x-1)^2}=14\). \[ \boxed{14} \]