The hidden quadratic — substitution from an indices/exponential equation

An equation contains the terms $2^{x+2}$ and $2^{x-1}$. Using the substitution $u = 2^x$, what do these two terms become?

• $4u$ and $\dfrac{u}{2}$

  Correct. $2^{x+2}=2^2\cdot2^x=4u$ and $2^{x-1}=2^x\div2=\dfrac{u}{2}$, applying the addition and subtraction laws respectively — the shift in the exponent becomes a constant FACTOR multiplying u, never a power of u.

• B$u^2$ and $u^{-1}$

  This is precisely the pair of errors a real WMA11 examiner report records for this exact type of question, translated from base 3 to base 2: "Common misconceptions included, $3^{x+2}=p^2$ and $3^{x-1}=p^{-1}$" (Jan 2023, Q5). Both treat an added or subtracted exponent as though it multiplied the exponent instead.

• C$2u$ and $-u$

  The shift constants (2 and -1) have been used as scalar multipliers of u directly, rather than as powers of the base 2. At x=0, u=1, and the real values are $2^2=4$ and $2^{-1}=0.5$ — neither matches 2×1=2 or -1×1=-1.

• D$u+2$ and $u-1$

  This applies the shift additively to u itself, as though substitution commuted with addition. $2^{x+2}$ is $2^x$ multiplied by $2^2$, not $2^x$ plus 2 — the two only coincide by accident for specific values of x, never in general.

Traps tested: Exponent shift mapped as a power not a multiple · Shift constant used directly instead of as a power of the base · Index shift applied additively to u

Solve $5^{2x} - 6 \times 5^x + 5 = 0$ for $x$.

• $x = 0$ or $x = 1$

  Correct. With $u=5^x$: $u^2-6u+5=(u-1)(u-5)=0$, so $u=1$ or $u=5$. Both are positive, so both survive: $5^x=1=5^0\Rightarrow x=0$, and $5^x=5=5^1\Rightarrow x=1$. $x=0$ is a genuine solution — $5^0=1$ is exact, not an approximation to be suspicious of.

• B$u = 1$ or $u = 5$

  This is the substituted variable's value, not x's. The question asks to solve for x; matching u=1 to x=0 and u=5 to x=1 is the step this response is missing.

• C$x = 1$ only — $u=1$ can be discarded, since $5^0=1$ doesn't look like a genuine solution

  u=1 is not a value to be suspicious of; it is a completely ordinary positive value of $5^x$, just as valid as u=5. Only a NEGATIVE value of u gets rejected, because $5^x$ can never be negative — u=1 fails no such test, and x=0 is a real solution to the original equation.

• DNo real solutions — solving $5x^2-6x+5=0$ directly by the quadratic formula gives a negative discriminant

  This treats the original equation as though it were already quadratic IN x, with x itself playing the role of the squared variable. It is quadratic in $5^x$, not in x — the exponent, not x itself, is what appears squared. $5x^2-6x+5=0$ is a different equation entirely, and its negative discriminant says nothing about the real one.

Traps tested: Stops at the substituted variable · Valid root wrongly discarded · Treats the original equation as already quadratic in x

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. (Verbatim WMA11 instruction, verified attached to real substitution-style questions; the equation below is VERIDIAN-original, not a reproduction.) Solve $3 \times 4^x - 11 \times 2^x - 4 = 0$.

Which response is certain to earn full marks?

• $(3u+1)(u-4)=0$ where $u=2^x$, giving $u=4$ (so $x=2$) and $u=-\dfrac{1}{3}$ (rejected, since $2^x>0$)

  Correct, and complete: the substitution is shown, the quadratic is solved by a named method, the invalid root is rejected with a reason, and the surviving root is converted all the way back to x. Nothing is left implicit.

• B$x = 2$, with no working shown

  A correct value with no visible method. The general marking guidance is explicit that answers without working may not gain full credit, and the printed instruction on this question removes the usual benefit of the doubt entirely.

• C"By calculator, $x=2$"

  This is the exact response the instruction exists to exclude. A real examiner report on this question type records that candidates who found their solution directly from a calculator, without showing a method, could have been penalised.

• DFull working shown up to $u=4$ or $u=-\dfrac{1}{3}$, with the response left there

  Every method mark up to this point is genuinely earned — the working is real and correctly shown. But the question asks to solve for x, and this response, however carefully written, never states x=2 or addresses the rejected root: showing your working thoroughly is not the same discipline as finishing the question, and this is the trap the lesson is named for, wearing a fully-worked-looking disguise.

Traps tested: Answer without working · Calculator cited as the method · Stops at the substituted variable

Solving a hidden-quadratic equation using the substitution $u=2^x$, you correctly reach $4u^2+4u-3=0$ and then $u=\dfrac{1}{2}$ or $u=-\dfrac{3}{2}$. What is the complete, correctly-justified solution for $x$?

• $x=-1$ only — $u=\dfrac{1}{2}=2^{-1}$ gives $x=-1$; $u=-\dfrac{3}{2}$ is rejected because $2^x$ is never negative

  Correct. Only one of the two u-values is a possible value of $2^x$ at all; the other is discarded before any attempt is made to convert it, and the surviving value is matched to its exponent directly ($\frac12=2^{-1}$), needing no logarithm.

• B$x=-1$ or $x=-\dfrac{3}{2}$

  The values of u have been reported directly as though they were already the values of x, with no rejection and no conversion — the two errors this lesson names combined into a single response.

• C$x=-1$ or $x=1$

  This assumes a negative value of u can be 'fixed' by flipping its sign and treating the result as a value of x. There is no law of indices that does this — a negative u is not a solution with the wrong sign, it is not a solution at all, and it is rejected, not corrected.

• DNo solution exists, since one of the two u-values is negative

  One invalid root does not disqualify the other. $u=\frac12$ is a genuine, positive value of $2^x$ and converts to a real solution, $x=-1$ — the correct response discards the bad root and keeps the good one, not both.

Traps tested: Stops at the substituted variable · Sign of rejected root flipped instead of rejected · Valid root abandoned because the other root failed