CAT 2005 | Question: 26

Question

Let $x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-}}}} \dots$ to infinity. Then $x$ equals

• A. $3$
• B. $\frac{\sqrt{13} -1}{2}$
• C. $\frac{\sqrt{13} +1}{2}$
• D. $\sqrt{13}$

Answer 1

Let
\[
x = \sqrt{4 + \sqrt{4 - \sqrt{4 + \cdots}}}
\]
\[
x = \sqrt{4 + y}, \quad y = \sqrt{4 - x}
\]

Squaring both:
\[
x^2 = 4 + y \Rightarrow y = x^2 - 4
\]
\[
y^2 = 4 - x
\]

Substitute:
\[
(x^2 - 4)^2 = 4 - x
\]

Expanding:
\[
x^4 - 8x^2 + 16 = 4 - x
\]
\[
x^4 - 8x^2 + x + 12 = 0
\]

So the valid root is: - put the options in the equation!
\[
x = \frac{\sqrt{13} - 1}{2}
\]

\[
\boxed{\text{Option B}}
\]