Moderate -0.3 This is a standard substitution problem where letting u = x^(1/3) transforms it into the quadratic u² - u - 6 = 0, which factors easily to (u-3)(u+2) = 0. While it requires recognizing the substitution pattern and handling fractional indices, it's a routine C1 exercise with straightforward steps, making it slightly easier than average.
Use a substitution to obtain a quadratic, or factorise into 2 brackets each containing \(x^{\frac{1}{3}}\). No marks if whole equation cubed/rooted etc. No marks if straight to quadratic formula with no evidence of substitution at start and no cube rooting/cubing at end
\(k^2-k-6=0\)
M1dep
Attempt to solve resulting three-term quadratic – see guidance in appendix 1
\((k-3)(k+2)=0\)
\(k=3, k=-2\)
A1
Correct values of \(k\). Spotted solutions: If M0 DMO or M1 DM0: SC B1 \(x=27\) www; SC B1 \(x=-8\) www
\(x=3^3, x=-2^3\)
M1
Attempt to cube at least one value
\(x=27, x=-8\)
A1
Final answers correct ISW (Can then get 5/5 if both found www and exactly two solutions justified)
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k=x^{\frac{1}{3}}$ | M1* | Use a substitution to obtain a quadratic, or factorise into 2 brackets each containing $x^{\frac{1}{3}}$. **No marks** if whole equation cubed/rooted etc. **No marks** if straight to quadratic formula with no evidence of substitution at start and no cube rooting/cubing at end |
| $k^2-k-6=0$ | M1dep | Attempt to solve resulting three-term quadratic – see guidance in appendix 1 |
| $(k-3)(k+2)=0$ | | |
| $k=3, k=-2$ | A1 | Correct values of $k$. **Spotted solutions:** If M0 DMO or M1 DM0: SC B1 $x=27$ www; SC B1 $x=-8$ www |
| $x=3^3, x=-2^3$ | M1 | Attempt to cube at least one value |
| $x=27, x=-8$ | A1 | Final answers correct **ISW** (Can then get 5/5 if both found www and exactly two solutions justified) |
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