Stand Up & Math · Teacher Answers
The Quadratic in Disguise
Not for distribution · Hidden quadratics · Full worked solutions
This is deliberately beyond "solve routine equations with indices" — the whole point is a technique (recognizing disguised quadratic structure via substitution) that doesn't exist anywhere in the standard curriculum. The student page gives no hints toward substitution, the rejected root, or clearing the fraction in Phase 4 — those are exactly the moments meant to be discovered, not read. Each phase below has a hint ladder: give at most one hint at a time, wait, and only escalate if a group is genuinely stuck rather than just moving slowly. A group that struggles for a while before the insight lands has had a better experience than one that's walked through it.
"This equation — 4 to the x, minus 5 times 2 to the x, plus 4, equals 0 — doesn't look like anything you already know how to solve. But it's wearing a disguise. Can your group unmask it?"
Write the equation on the board rather than only saying it aloud. Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.
Rewriting \(4^x\) in terms of \(2^x\)
Letting \(y=2^x\), the equation \(4^x-5(2^x)+4=0\) becomes:
This is an ordinary quadratic equation in \(y\).
The whole activity hinges on Task 1a. Groups that struggle here usually haven't connected "power of a power" to this direction of use — they've only ever applied it forwards (simplify \((2^3)^2\)) not backwards (recognize \(4^x\) as \((2^x)^2\)). Use the hint ladder above rather than pointing this out directly.
Require: correct rewrite of \(4^x\), correct full substitution with no \(x\) remaining, and the equation identified as quadratic.
Solving the hidden quadratic
Require: both values of \(y\) found, both correctly translated to \(x\), and both verified in the original equation — the checking step matters, since Phase 3 depends on them being comfortable with this habit.
When a root has to be rejected
Since \(y=3^x\) and an exponential is always positive, \(y=-5\) is impossible — reject it. Only \(y=9\) is valid.
This is the crux of the phase. Some groups will happily write "\(x=?\)" for \(3^x=-5\) without noticing anything is wrong — worth letting them try to solve it and hit the wall themselves (there's no real \(x\) with \(3^x\) negative) rather than pointing it out first.
Require: both values of \(y\) found, \(y=-5\) explicitly rejected with a stated reason (not just silently dropped), and the correct final answer \(x=2\).
Proving no solution exists
Negative discriminant — no real solutions for \(y\), which means no real \(x\) satisfies the original equation.
Accept the discriminant argument fully — proving impossibility via "the quadratic has no real roots" is a completely valid and sufficient proof at this level. The AM–GM route in the challenge extension is a bonus, not a requirement.
Require: correct quadratic formed, discriminant correctly computed and identified as negative, and an explicit conclusion connecting this back to the original equation having no real solution.
For any \(y>0\), \(y+\dfrac1y \ge 2\), with equality only when \(y=1\). This is the AM–GM inequality applied to \(y\) and \(\frac1y\) (their product is always 1, so their sum is minimized when they're equal). Since \(y=2^x>0\) always, the smallest possible value of \(2^x+2^{-x}\) is 2 — it can never reach 1.
This is genuinely olympiad-level reasoning. Most groups won't reach it unprompted — that's fine, it's a stretch goal, and the student page deliberately gives no hint toward it. If a group is engaged but stuck, one hint: "what's the smallest possible value of \(y+\frac1y\) for \(y>0\)?" If they reach it independently, flag explicitly that they've just used a real inequality technique (AM–GM) that goes well beyond this course.
1. Have a group present Phase 1's key insight — rewriting \(4^x\) as \((2^x)^2\) — since this is the single move the entire activity depends on.
2. Ask the class: "when might a substitution give you a value that turns out to be impossible?" before revealing Phase 3's rejected root, to see if anyone predicts it.
3. For Phase 4, ask for a show of hands on whether they think a solution exists before revealing the discriminant is negative — a good moment for genuine mathematical suspense.
4. If time allows, connect this technique's name: this substitution trick generalizes to any equation of the form \(a^{2x}+b\cdot a^x+c=0\), and shows up again properly when they meet exponential and logarithmic equations in later years.