How to Solve Infinite Nested Roots — Tricks & Step-by-Step Methods
Simple, reliable approaches for nested radicals and roots that often appear in competitive exams and problem sets.
Infinite nested roots look scary but are usually straightforward if you follow the same three steps:
- Set the whole expression equal to a variable, usually
y. - Use the structure of the nest to write an equation in
y(square/cube both sides as needed). - Solve the resulting equation (choose the non-negative root consistent with the radical).
Common Patterns and Formulas
- Square-root additive (positive):
\[ y=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}\quad\Longrightarrow\quad y=\sqrt{x+y} \] Solve: \(y^2-y-x=0\) → \(y=\dfrac{1+\sqrt{1+4x}}{2}\). - Square-root additive (negative):
\[ y=\sqrt{x-\sqrt{x-\sqrt{x-\dots}}}\quad\Longrightarrow\quad y=\sqrt{x-y} \] Solve: \(y^2+y-x=0\) → \(y=\dfrac{-1+\sqrt{1+4x}}{2}\). - n-th root multiplicative pattern (common for cube examples):
Expression like \(y=\sqrt[n]{x\sqrt[n]{x\sqrt[n]{x\cdots}}}\) leads to \(y^n = x\,y\) → \(y^{\,n-1}=x\) → \(\displaystyle y = x^{1/(n-1)}\) (choose real positive where appropriate). - General n-th root additive:
\(y=\sqrt[n]{x+y}\) → \(y^n-y-x=0\). For \(n\ge3\) you usually solve numerically or analyze existence/uniqueness of the positive root.
Worked Examples (from the page you shared)
Example 1 — Positive square root
\[ \sqrt{30+\sqrt{30+\sqrt{30+\dots}}} \]Step 2: Square both sides: \(y^2=30+y\) → \(y^2-y-30=0\).
Step 3: Solve quadratic: \[ y=\frac{1\pm\sqrt{1+120}}{2}=\frac{1\pm\sqrt{121}}{2}=\frac{1\pm11}{2}. \] Only the positive root makes sense for nested radicals: \(y=\dfrac{1+11}{2}=6\).
Answer: 6
Quick trick used in many sources: Factor 30 as \(6\times5\) — the two numbers 6 and 5 appear as the two quadratic roots; for a + series take the larger root (6).
Example 2 — Negative square root
\[ \sqrt{30-\sqrt{30-\sqrt{30-\dots}}} \]Answer: 5
Trick: factor 30 = 6×5 again — for a − series the smaller positive root (5) is the correct value.
Example 3 — Nested √7 (additive)
\[ \sqrt{7+\sqrt{7+\sqrt{7+\dots}}} \]Answer: \(\dfrac{1+\sqrt{29}}{2}\)
Example 4 — Cube-root multiplicative
\[ \sqrt[3]{23\sqrt[3]{23\sqrt[3]{23\cdots}}} \]Answer: \(\sqrt{23}\)
Why we pick the positive root
All nested expressions built from real square roots (or n-th root with odd n) are non-negative by construction. So after solving the polynomial equation, we discard negative roots unless the expression could allow them (rare). Always check domain constraints: e.g., inside a square root the value must be ≥ 0.
Common pitfalls
- For expressions like \(y=\sqrt{x+y}\), do not treat both quadratic roots as valid — test which root satisfies the original (non-squared) equation.
- When manipulating, squaring can introduce extraneous roots; verify by substitution back into the original nested radical form.
- For generalized \(y=\sqrt[n]{x+y}\) with \(n\ge3\), algebraic closed-forms may not exist; numeric methods or monotonicity arguments are used to find the positive real root.
Short proofs & existence notes (click to expand)
For many additive nested radicals \(y=\sqrt{x+y}\), you can show monotonic sequence convergence: define sequence \(y_1=\sqrt{x}\), \(y_{k+1}=\sqrt{x+y_k}\). If this sequence is increasing and bounded above, it converges to the positive root of \(y^2-y-x=0\).
Quick Reference Table
| Form | Equation | Solution (select) |
|---|---|---|
| \( \sqrt{x+\sqrt{x+\dots}} \) | \(y^2-y-x=0\) | \(y=\dfrac{1+\sqrt{1+4x}}{2}\) |
| \( \sqrt{x-\sqrt{x-\dots}} \) | \(y^2+y-x=0\) | \(y=\dfrac{-1+\sqrt{1+4x}}{2}\) |
| \( \sqrt[n]{x\sqrt[n]{x\cdots}} \) | \(y^n = x\,y\) | \(y = x^{1/(n-1)}\) (positive) |
| \( \sqrt[n]{x+\sqrt[n]{x+\dots}} \) | \(y^n-y-x=0\) | Numeric solution (positive root) |
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