Standard +0.3 This is a quadratic-in-disguise problem requiring substitution u = x³, solving 8u² + 7u - 1 = 0 by factorization or formula, then finding x from u. While it requires recognizing the substitution pattern and cube roots, it's a standard C1 technique with straightforward execution, making it slightly easier than average.
Guidance: Use a substitution to obtain a quadratic or factorise into 2 brackets each containing \(x^3\)
Correct method to solve a quadratic
Both values of \(k\) correct
Spotted solutions: If M0 DMO or M1 DM0 SC B1 \(x = -1\) www; SC B1 \(x = \frac{1}{2}\) www (Can then get 5/5 if both found www and exactly two solutions justified)
Additional notes: No marks if whole equation cube rooted etc. No marks if straight to formula with no evidence of substitution at start and no cube rooting/cubing at end.
**Answer:** $k = x^3$; $8k^3 + 7k - 1 = 0$; $(8k-1)(k+1) = 0$; $k = \frac{1}{8}, k = -1$; $x = \frac{1}{2}, x = -1$
**Marks:** M1* | DM1* | A1 | M1 | A1 [5]
**Guidance:** Use a substitution to obtain a quadratic or factorise into 2 brackets each containing $x^3$ | Correct method to solve a quadratic | Both values of $k$ correct | Attempt to cube root at least one value to obtain $x$ | Both values of $x$ correct and no other values
**Spotted solutions:** If M0 DMO or M1 DM0 SC B1 $x = -1$ www; SC B1 $x = \frac{1}{2}$ www (Can then get 5/5 if both found www and exactly two solutions justified)
**Additional notes:** No marks if whole equation cube rooted etc. No marks if straight to formula with no evidence of substitution at start and no cube rooting/cubing at end.
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