If \(\rm x=\frac{6}{7-\frac{6}{7-\frac{6}{7-\frac{6}{7-x}}}}\); x > 1, then what is the value of x2 - 3x + 2 equal to ?

 Shortcut Trick

\(\rm x=\frac{6}{7-\frac{6}{7-\frac{6}{7-\frac{6}{7-x}}}}\)

Clearly we can see, at x = 6,

LHS = RHS

Therefore,

x2 - 3x + 2 = 6² - 3 × 6 + 2

x2 - 3x + 2 = 36 - 18 + 2

∴ x2 - 3x + 2 = 20

Alternate MethodGiven:

\( x = \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - x}}}}, x > 1 \)

Calculate \( x^2 - 3x + 2 \)

Concept Used:

Recurrence and simplification of nested fractions followed by algebraic simplification.

Calculation:

Let's solve the recurrence:

⇒ Let \( y = 7 - x \).

⇒ Then \( x = \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - \frac{6}{7 - x}}}} \)

⇒ Substitute \( y \) back in, \( x = \frac{6}{y} \) where \( y = 7 - x \)

⇒ \( x = \frac{6}{7 - x} \)

⇒ \( x(7 - x) = 6 \)

⇒ \( 7x - x^2 = 6 \)

⇒ \( x^2 - 7x + 6 = 0 \)

⇒ \( (x - 6)(x - 1) = 0 \)

⇒ \( x = 6 \) or \( x = 1 \) (Since \( x > 1 \), \( x = 6 \))

Now, calculate \( x^2 - 3x + 2 \):

⇒ Substitute \( x = 6 \) into \( x^2 - 3x + 2 \),

⇒ \( 6^2 - 3 \times 6 + 2 \)

⇒ \( 36 - 18 + 2 \)

⇒ \( 20 \)

∴ \( x^2 - 3x + 2 = 20 \).