3 Ways of Evaluating Nested Square Roots

Nested square roots or nested radical problems are quite interesting to solve. The key skill for this question is to understand how the students can handle “…”. This enables us to set up a quadratic equation to evaluate its exact value using the quadratic formula,
$$x= \frac{-b \ \pm \sqrt{b^2-4ac}}{2a}$$.
Let’s look at the following examples for finding the nested square roots.

Example 1

Evaluate \( \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}} \)

\( \begin{aligned} \require{AMSsymbols} \require{color}
\text{Let } x &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} &\color{green} {(1)} \\
x^2 &= 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &\color{green}{\text{square both sides}} \\
x^2-2 &= \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &\color{green}{\text{move 2 to the left}} \\
x^2-2 &= x &\color{green}{\text{replace the nested square root by } (1)} \\
x^2-x-2 &= 0 &\color{green}{\text{form a quadratic equation}} \\
x &= \frac{1 \pm \sqrt{(-1)^2-4 \times 1 \times (-2)}}{2} &\color{green}{\text{apply into quadratic formula}} \\
x &= \frac{1 \pm \sqrt{9}}{2} \\
x &= \frac{1 \pm 3}{2} \\
x &= \frac{1 + 3}{2} \text{ or } \frac{1-3}{2} \\
x &= 2 \text{ or } -1 \\
x &= 2 &\color{green} {\text{ as } \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} > 0} \\
\therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} &= 2
\end{aligned} \)

Example 2

Evaluate \( \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} \)

\( \begin{aligned} \require{AMSsymbols} \require{color}
\text{Let } x &= \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\ldots}}}} &\color{green} {(1)} \\
x^2 &= 2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}} &\color{green}{\text{square both sides}} \\
x^2-2 &= -\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}} &\color{green}{\text{move 2 to the left}} \\
x^2-2 &= -x &\color{green}{\text{replace the nested square root by } (1)} \\
x^2 + x-2 &= 0 &\color{green}{\text{form a quadratic equation}} \\
x &= \frac{-1 \pm \sqrt{1^2-4 \times 1 \times (-2)}}{2} &\color{green}{\text{apply into quadratic formula}} \\
x &= \frac{-1 \pm \sqrt{9}}{2} \\
x &= \frac{-1 \pm 3}{2} \\
x &= \frac{-1 + 3}{2} \text{ or } \frac{-1-3}{2} \\
x &= 1 \text{ or } -2 \\
x &= 1 &\color{green} {\text{ as } \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} > 0} \\
\therefore \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} &= 1
\end{aligned} \)

Example 3

Evaluate \( \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}} \) using the double angle trigonometric property.

\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\cos\frac{\pi}{4} &= \frac{\sqrt{2}}{2} &\color{green}{(1)} \\
2 \cos^2\frac{\pi}{8}-1 &= \cos\frac{\pi}{4} &\color{green}{\text{apply the double-angle formula}} \\
2 \cos^2\frac{\pi}{8}-1 &= \frac{\sqrt{2}}{2} &\color{green}{\text{from (1)}} \\
2 \cos^2\frac{\pi}{8} &= 1 + \frac{\sqrt{2}}{2} &\color{green}{\text{move -1 to the right hand side}} \\
2 \cos^2\frac{\pi}{8} &= \frac{2 + \sqrt{2}}{2} &\color{green}{\text{single fraction}} \\
\cos^2\frac{\pi}{8} &= \frac{2 + \sqrt{2}}{4} &\color{green}{\text{divide both sides by 2}} \\
\cos\frac{\pi}{8} &= \frac{\sqrt{2 + \sqrt{2}}}{2} &\color{green}{(2)} \\
\cos\frac{\pi}{16} &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} &\color{green}{(3)} \\
\ldots \\
\cos\frac{\pi}{\infty} &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} &\color{green}{(4)} \\
\cos 0 &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} \\
1 &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}}{2} \\
\therefore \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} &= 2
\end{aligned} \)

 

Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume




Comments

  1. George Plousos

    What would you say about that?
    √2±(√2±(√2±(√2± … ±(√2)…)))=?
    For any combination of signs.

    1. iitutor Post author

      It is a mixture of example 1 and 2. Which means the answer is either 2 or 0.

Your email address will not be published. Required fields are marked *