Challenging +1.2 This question requires recognizing that a substitution can simplify a quartic equation to a quadratic, then solving the quadratic and back-substituting. While the substitution is given (removing the hardest step), students must manipulate the quartic algebraically to express it in terms of u, solve the resulting quadratic, then solve two separate quadratic equations. This multi-stage process with algebraic manipulation is moderately above average difficulty but follows a clear pathway once the substitution is applied.
7 In this question you must show detailed reasoning.
Use the substitution \(u = 6 x ^ { 2 } + x\) to solve the equation \(36 x ^ { 4 } + 12 x ^ { 3 } + 7 x ^ { 2 } + x - 2 = 0\).
\(u^2 = 36x^4 + 12x^3 + x^2\); So \(36x^4 + 12x^3 + 7x^2 + x - 2 = u^2 + 6x^2 + x - 2\)
M1
\((36x^4 + 12x^3 + 7x^2 + x - 2) \div (6x^2 + x) = 6x^2 + x + 1\) rem \(-2\); \(((6x^2+x)(6x^2+x+1) = 2)\); M1 for attempt \((6x^2+x)^2\) or attempt \(\div\) LHS by \((6x^2+x)\), at least 2 terms correct, or obtain any correct eqn in terms of \(x\) and \(u\)
Eqn reduces to \(u^2 + u - 2 = 0\); \(u = -2\) or \(1\)
A1, A1
\(u(u+1) = 2\); BC
\(6x^2 + x = -2\) has no roots because \(\Delta = 1 - 48 < 0\)
B1
Must see correct calc'n for \(\Delta\) and "\(< 0\)" for their quadratic equation; or \(x = \frac{-1\pm\sqrt{47}i}{12}\) given instead of "no roots" etc
\(6x^2 + x = 1\) has roots \(x = \frac{1}{3}\) or \(-\frac{1}{2}\)
A1
BC Ignore any answers from \(u = -2\); SC If M1 gained but incorrect or inadequate method & correct answers: M1A0A0B0A1; Otherwise correct ans without any correct working: no marks. (Because DR)
# Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u^2 = 36x^4 + 12x^3 + x^2$; So $36x^4 + 12x^3 + 7x^2 + x - 2 = u^2 + 6x^2 + x - 2$ | M1 | $(36x^4 + 12x^3 + 7x^2 + x - 2) \div (6x^2 + x) = 6x^2 + x + 1$ rem $-2$; $((6x^2+x)(6x^2+x+1) = 2)$; M1 for attempt $(6x^2+x)^2$ or attempt $\div$ LHS by $(6x^2+x)$, at least 2 terms correct, or obtain any correct eqn in terms of $x$ and $u$ |
| Eqn reduces to $u^2 + u - 2 = 0$; $u = -2$ or $1$ | A1, A1 | $u(u+1) = 2$; **BC** |
| $6x^2 + x = -2$ has no roots because $\Delta = 1 - 48 < 0$ | B1 | Must see correct calc'n for $\Delta$ and "$< 0$" for their quadratic equation; or $x = \frac{-1\pm\sqrt{47}i}{12}$ given instead of "no roots" etc |
| $6x^2 + x = 1$ has roots $x = \frac{1}{3}$ or $-\frac{1}{2}$ | A1 | **BC** Ignore any answers from $u = -2$; SC If M1 gained but incorrect or inadequate method & correct answers: M1A0A0B0A1; Otherwise correct ans without any correct working: no marks. (Because **DR**) |
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7 In this question you must show detailed reasoning.\\
Use the substitution $u = 6 x ^ { 2 } + x$ to solve the equation $36 x ^ { 4 } + 12 x ^ { 3 } + 7 x ^ { 2 } + x - 2 = 0$.
\hfill \mbox{\textit{OCR H240/02 2019 Q7 [5]}}