Stand Up & Math · Teacher Answers
Radicals All the Way Down
Not for distribution · Denesting radicals · Full worked solutions
Nothing here is new content — every step is Section A expansion/simplification skill, aimed at a job students have never used it for: running the process backward. The mathematical payoff (a golden-ratio reveal hiding inside an infinite radical) is real, not a gimmick, so it's worth protecting the surprise — avoid confirming where Phase 4 is headed until they get there themselves.
"Here's an expression: √(3+2√2). Can your group find its exact value — no calculator, no decimals? You already have every skill you need to do this. You just don't know it yet."
Do not confirm or deny any early guesses. Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.
Expand and look for a pattern
Comparing each result to its starting pair \((m,n)\): the whole-number part is always \(m+n\), and the number under the new root is always \(mn\) — with the surd's coefficient always 2.
The third example, \((\sqrt2+\sqrt3)^2\), is the one that makes the pattern click for most groups — it's the only one of the three where both starting terms are already surds, which makes \(m\) and \(n\) visually obvious. If a group is stuck, suggest they revisit that one first.
\(\sqrt{3+2\sqrt2} = 1+\sqrt2\), directly from the Task 1a result. Most groups won't notice this link on their own — it's worth pointing out explicitly once they reach it, since realizing they already solved the "impossible" opening problem is a genuine payoff moment.
Require: all three expansions correct, and the general pattern \((m+n)+2\sqrt{mn}\) stated in some form (exact notation doesn't matter, the relationship does).
Reverse-engineer the pattern
By inspection, \(m=2,\ n=3\) (sum 5, product 6).
Require: both equations correctly set up, correct pair found, and the final answer written in \(\sqrt{m}+\sqrt{n}\) form (not just "m=2, n=3" left unfinished).
This "two numbers with a given sum and product" move is exactly the technique they'll meet again later this year when factoring quadratics by inspection. Naming that connection now plants a useful seed.
When the pattern doesn't quite fit
So \(7+4\sqrt3\) becomes \(7+2\sqrt{12}\), which now matches the Phase 2 pattern.
Need \(m+n=7,\ mn=12\): \(m=3,\ n=4\).
Groups that skip Task 3a and try to force \(m+n=7,\ mn=3\) directly will get stuck (no whole-number solution) — that failure is useful, not a problem. Let them sit with it briefly before nudging them back to the coefficient rewrite. Also watch that they simplify \(\sqrt4\) to 2 rather than leaving the answer as \(\sqrt3+\sqrt4\).
Accept any version that captures: (1) check whether the surd's coefficient is already 2 — if not, rewrite it as \(2\sqrt{\text{something}}\); (2) match the whole-number part to \(m+n\) and the number under the new root to \(mn\); (3) solve for \(m,n\); (4) write \(\sqrt{m}+\sqrt{n}\) and simplify any resulting surds.
Require: correct coefficient rewrite, correct final answer \(2+\sqrt3\), and a genuine attempt at the general method in Task 3d — this is the synthesis step, worth pushing for real articulation rather than "just do what we did before."
A radical that never ends
The expression under the very first root — \(1+\sqrt{1+\sqrt{1+\cdots}}\) — is exactly the same infinite pattern as \(x\) itself, since it continues identically forever. So the "\(1+\cdots\)" under the first root can be replaced with \(x\), giving \(x=\sqrt{1+x}\).
This self-similarity argument is the conceptual heart of the phase and is genuinely unfamiliar reasoning — expect groups to need a moment with it. A useful nudge: "if you deleted the outermost square root and the '1+', would what's left look any different from x?"
The negative root \(\frac{1-\sqrt5}{2}\approx-0.618\) is rejected since \(x\) is built entirely from square roots and must be positive.
This is the golden ratio, \(\varphi\). Accept any reference students bring up (art, architecture, the Parthenon, nautilus shells, Fibonacci numbers) — the point is recognition, not a specific correct source.
Require: correctly formed quadratic, both roots found via the formula, correct root selected with a reason given (not just "the positive one"), and the decimal value stated.
Rejecting the negative root, \(x=2\) exactly — a clean integer emerging from an infinite radical.
Groups that finish this and ask "does this work for any starting number?" have found real mathematical territory — the general result is \(x=\frac{1+\sqrt{1+4k}}{2}\) for a nest of \(k\)'s. Confirm it works but let them test it themselves rather than just handing it over.
1. Ask the whole class to answer the opening challenge together, out loud, before revealing it was already solved back in Phase 1.
2. Have a group walk through the coefficient-rewriting trick from Phase 3 — this is the step most likely to be forgotten later, worth reinforcing.
3. Build up to Phase 4 as a "reveal": ask what number they got, then ask if anyone knows its name, before confirming it's the golden ratio.
4. If time allows, mention that this exact self-similarity trick (writing an infinite structure in terms of itself) is the same idea behind recursion in computer science and behind continued fractions.