Self-Similar Expressions: Practice

Name the tower $x$. The part under the first radical is $2$ plus a copy of the whole tower, so $x = \sqrt{2 + x}$. Square and factor.

$x^2 - x - 2 = 0 \quad\Longrightarrow\quad (x - 2)(x + 1) = 0$

The candidates are $2$ and $-1$, but a square root is never negative, so the value is $x = 2$.

Under the outermost radical sits $6$ plus another copy of the entire tower, which is the same $x$.

$x = \sqrt{6 + x}$

The copy inside must be identical to the whole, so it is $6 + x$ under a single square root.

Let $x = 0.\overline{3}$. Multiplying by $10$ exposes a copy of $x$: $10x = 3.\overline{3} = 3 + x$.

$9x = 3 \quad\Longrightarrow\quad x = \frac{1}{3}$

So $0.\overline{3} = \tfrac{1}{3}$.

The repeating block $3$ is one digit long, so slide the digits one place with a factor of $10$.

$10x = 3.\overline{3} = 3 + x$

The tail after the decimal point is the original $x$ again.

Set $x = \sqrt{12 + x}$, then square and factor.

$x^2 - x - 12 = 0 \quad\Longrightarrow\quad (x - 4)(x + 3) = 0$

Discard the negative candidate, so the value is $x = 4$.

The symbol $\sqrt{\phantom{x}}$ returns the principal root, which is never negative, and the tower is built from positive parts.

$x = \sqrt{2 + x} \ \ge \ 0$

So a negative candidate cannot be the value of the expression.

Set $x = \sqrt{20 + x}$, then square and factor.

$x^2 - x - 20 = 0 \quad\Longrightarrow\quad (x - 5)(x + 4) = 0$

The value is the nonnegative candidate, $x = 5$.

Self-similarity means the whole expression reappears inside itself, so naming the whole $x$ lets you replace the inner copy by $x$.

$x = \sqrt{2 + x}$

That single step turns an infinite process into a finite equation.

Multiply every term by $x$ to clear the fraction.

$x \cdot x = x \cdot 1 + x \cdot \tfrac{1}{x} \quad\Longrightarrow\quad x^2 = x + 1$

Moving everything to one side gives $x^2 - x - 1 = 0$.

Name it $x$. Under the outer radical sits $2$ times a copy of the whole tower, so $x = \sqrt{2x}$. Square and factor.

$x^2 - 2x = 0 \quad\Longrightarrow\quad x(x - 2) = 0$

The tower is larger than $1$, so $x = 2$.

The repeating block $45$ is two digits long, so slide the digits two places with a factor of $100$.

$100x = 45.\overline{45} = 45 + x$

The tail after the decimal point is the original $x$ again.

Name it $x$. Under the outer radical sits $3$ times a copy of the whole tower, so $x = \sqrt{3x}$. Square and factor.

$x^2 - 3x = 0 \quad\Longrightarrow\quad x(x - 3) = 0$

The tower exceeds $1$, so $x = 3$.