A general problem-solving technique video — not a specific exam question.
Problem
Find the value of $\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}$.
Key insight. Because the nested pattern repeats forever, the expression inside the outermost square root is identical to the whole expression itself. Naming the whole thing $x$ turns the infinite nesting into a single equation, $x=\sqrt{12+x}$ — no need to ever “unroll” more than one layer.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Let $x$ equal the entire infinite expression. Since the pattern repeats indefinitely, the expression under the outermost square root is just $12+x$ again. Square both sides to remove the root, solve the resulting quadratic, and reject any root that can’t represent a square root (which must be non-negative).
Solution
Step 1 — Name the expression x
$$x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}$$
Step 2 — Recognise the self-similar structure
Since the nesting continues forever, the part under the first square root is the same infinite expression:
$$x = \sqrt{12+x}$$
Step 3 — Square both sides
$$x^2 = 12+x$$
$$x^2-x-12=0$$
Step 4 — Factor the quadratic
Splitting the middle term: $x^2-4x+3x-12=0$
$$x(x-4)+3(x-4)=0 \implies (x-4)(x+3)=0$$
Step 5 — Solve and select the valid root
$$x=4 \quad \text{or} \quad x=-3$$
Since $x$ is defined as a square root (of a positive quantity), it must be non-negative. Reject $x=-3$.
Answer
$$4$$
Common mistakes
- Trying to compute the expression by nesting a few layers deep numerically. This approach can approximate the answer but never proves it exactly — recognising the self-similarity and setting up the equation $x=\sqrt{12+x}$ is what gives the exact value directly.
- Keeping both roots of the quadratic as valid answers. Since $x$ represents a square root, it must be non-negative by definition — the negative root $x=-3$ is a valid solution of the quadratic but not of the original nested radical.
Practise next
- Find the value of $\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}$, using the same self-similarity and quadratic approach.
Show answer
$5$. Call the value $x$. The expression under the first root is $20$ plus the whole expression again, so $x=\sqrt{20+x}$.
Squaring, $x^2-x-20=0$, which factors as $(x-5)(x+4)=0$.
A square root is never negative, so $x=5$. The self-similarity is the whole idea: removing one layer leaves the same infinite expression.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.