The "hidden" higher-degree equation, reached via differentiation

The curve $y = x^3 + \dfrac{2}{x}$ ($x \neq 0$) has $\dfrac{dy}{dx} = 5$ at certain points. Find all values of $x$ for which this is true.

• A$x = \pm1$

  This is the wrong pair of roots for this equation — worth checking directly rather than trusting on sight: at $x=1$, $\frac{dy}{dx}=3(1)-\frac{2}{1}=1 \neq 5$. $x=\pm1$ would answer a similarly-shaped question with different coefficients, not this one.

• $x = \pm\sqrt2$

  Correct. $\frac{dy}{dx}=3x^2-\frac{2}{x^2}=5 \Rightarrow 3x^4-2=5x^2 \Rightarrow 3x^4-5x^2-2=0$. As a quadratic in $u=x^2$: $(3u+1)(u-2)=0$ [check: $3u^2-6u+u-2=3u^2-5u-2$ ✓], so $u=-\frac13$ or $u=2$. Reject $u=-\frac13$ since $x^2\geq0$; keep $u=2 \Rightarrow x=\pm\sqrt2$ — check: at $x=\sqrt2$, $3(2)-\frac{2}{2}=6-1=5$ ✓.

• C$x = \sqrt2$ only

  The valid value of $u$ is correctly found and correctly kept, but only the positive square root is taken. $x^2=2$ is satisfied by both $x=\sqrt2$ and $x=-\sqrt2$ — dropping the negative branch loses half the answer.

• D$x = -\dfrac{1}{\sqrt3}$ or $x = \pm\sqrt2$

  This attempts to convert BOTH values of u, including $u=-\frac13$ — but $x^2$ can never equal a negative number for real $x$, so that root has no real square root at all and must be rejected, not converted with a flipped sign.

Traps tested: Roots from a different equation · Negative square root branch dropped · Invalid negative u not rejected

A student, solving a differentiation question, reaches $6x^4 - 5x^2 - 6 = 0$ and writes: "$x^2 = \frac{3}{2}$ or $x^2 = -\frac23$, so $x = \frac32$." What is the single biggest error in this response?

• The surviving value, $x^2 = \frac32$, was never square-rooted — the final line should read $x = \pm\sqrt{\frac32}$, not $x = \frac32$

  Correct. Everything up to "$x^2 = \frac32$" is right (and $x^2=-\frac23$ is correctly implied to be rejected, since it's negative), but the response then reports the value of $x^2$ as though it were the value of $x$ — the exact documented trap this lesson is built around, confirmed on a real WMA11 script that reached $x^2=\frac14$ and stopped there in the same way.

• BThe quadratic $6u^2-5u-6=0$ was solved incorrectly

  Check it: $6u^2-5u-6=0$ factorises as $(3u+2)(2u-3)=0$ [check: $6u^2-9u+4u-6=6u^2-5u-6$ ✓], giving $u=-\frac23$ or $u=\frac32$ — both values in the response are correct. The quadratic-solving step is not where the error is.

• C$x^2 = -\frac23$ should not have been discarded — it has a valid real solution

  $x^2$ is never negative for real $x$, so $x^2=-\frac23$ genuinely has no real solution and is correctly discarded. This part of the response is right, not wrong.

• DThe original quartic should not have been treated as a quadratic in x² at all

  It should — $6x^4-5x^2-6=0$ has no x³ or x¹ term, which is exactly the signal that the substitution $u=x^2$ applies. Treating it as a quadratic in x² is the correct first move, not the error.

Traps tested: Blames the wrong step · Valid rejection treated as an error · Correct strategy flagged as wrong

A curve has a stationary point at $x = 2$, where $y = 7$. What is the equation of the NORMAL to the curve at this point?

• $x = 2$

  Correct. A stationary point has gradient 0, so the tangent there is horizontal ($y=7$). A line perpendicular to a horizontal line is vertical, and a vertical line through $(2,7)$ has equation $x=2$ — it cannot be written in the form $y=mx+c$ at all, since a vertical line has no defined gradient.

• B$y = 7$

  This is the equation of the TANGENT at this point, not the normal. The normal is perpendicular to the tangent — perpendicular to a horizontal line is vertical, not another horizontal line.

• CThe normal does not exist, since the perpendicular-gradient formula $m_1 m_2 = -1$ requires dividing by the tangent's gradient, which is 0

  The normal exists; the FORMULA breaks down, and those are different things. Where the algebraic route fails (division by zero), the geometric reasoning still works: perpendicular to a horizontal line is a vertical line, reached by reasoning about the picture rather than by the formula.

• D$y - 7 = 0(x - 2)$, i.e. $y = 7$

  This substitutes the TANGENT's gradient (0) into the point-gradient formula, which correctly reproduces the tangent — but the question asks for the normal, which needs the perpendicular direction, not the same one.

Traps tested: Tangent and normal confused · Formula failure treated as non existence · Tangent gradient used for the normal

Which of these equations is a "hidden quadratic in x²", solvable by substituting $u = x^2$?

• $4x^4 - 3x^2 + 1 = 0$

  Correct — every term is an even power of x (x⁴, x², and a constant, which is x⁰), so u = x² converts it directly into 4u² − 3u + 1 = 0, an ordinary three-term quadratic in u.

• B$4x^4 - 3x^3 + 1 = 0$

  The $x^3$ term breaks the pattern: substituting $u=x^2$ turns $x^4$ into $u^2$ and the constant into $u^0$, but $x^3$ is an ODD power and has no clean expression in u alone — it becomes $u^{1.5}$, not a whole-number power. This is not a hidden quadratic in x².

• C$4x^4 - 3x + 1 = 0$

  The $x^1$ term is an odd power, for the same reason as option B: $x$ does not become a whole-number power of $u=x^2$ (it would need $u^{0.5}$). An x¹ term anywhere rules out this particular substitution.

• D$4x^5 - 3x^2 + 1 = 0$

  $x^5$ is an odd power of $x$ (5 is odd), so it does not convert to a whole-number power of $u=x^2$ either — only EVEN powers of x (x², x⁴, x⁶, ...) do. A degree-5 equation cannot be a straightforward quadratic in x² no matter which of its other terms look promising.

Traps tested: Odd power term ignored · Even degree assumed from highest term alone