Standard +0.3 This is a standard substitution question where students let u = y^(1/4) to transform into 2u² - 7u + 3 = 0, then solve the quadratic and back-substitute. It's slightly above average difficulty due to the fractional powers and need to check validity of solutions, but follows a well-practiced technique with clear steps.
Use substitution to obtain quadratic or factorise into two brackets each containing \(y^{\frac{1}{4}}\)
\((2x-1)(x-3)=0\)
M1dep*
Correct method to solve resulting quadratic
\(x = \frac{1}{2},\ x = 3\)
A1
Both values correct
\(y = \left(\frac{1}{2}\right)^4,\ y = 3^4\)
M1dep*
Attempt to raise to the fourth power
\(y = \frac{1}{16},\ y = 81\)
A1
Correct final answers
No marks if whole equation raised to fourth power etc. No marks if straight to formula with no evidence of substitution at start and no raising to fourth power/fourth rooting at end. No marks if \(y^{\frac{1}{4}}=x\) and then \(2x - 7x^2 + 3 = 0\). Spotted solutions: If M0 DM0 or M1 DM0: SC B1 \(y=81\) www; SC B1 \(y=\frac{1}{16}\) www.
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $y^{\frac{1}{4}} = x$; $2x^2 - 7x + 3 = 0$ | M1* | Use substitution to obtain quadratic or factorise into two brackets each containing $y^{\frac{1}{4}}$ |
| $(2x-1)(x-3)=0$ | M1dep* | Correct method to solve resulting quadratic |
| $x = \frac{1}{2},\ x = 3$ | A1 | Both values correct |
| $y = \left(\frac{1}{2}\right)^4,\ y = 3^4$ | M1dep* | Attempt to raise to the fourth power |
| $y = \frac{1}{16},\ y = 81$ | A1 | Correct final answers |
**No marks** if whole equation raised to fourth power etc. **No marks** if straight to formula with no evidence of substitution at start and no raising to fourth power/fourth rooting at end. **No marks** if $y^{\frac{1}{4}}=x$ and then $2x - 7x^2 + 3 = 0$. **Spotted solutions:** If M0 DM0 or M1 DM0: SC B1 $y=81$ www; SC B1 $y=\frac{1}{16}$ www.
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