Standard +0.3 This is a standard Further Maths question on transformed roots requiring Vieta's formulas and algebraic manipulation. While it involves more steps than basic A-level questions, the technique is routine for FP1 students: find sum and product of new roots using α+β=-k and αβ=2k, then construct the new equation. The transformation is straightforward (reciprocal swap) with no novel insight required.
8 The quadratic equation \(x ^ { 2 } + k x + 2 k = 0\), where \(k\) is a non-zero constant, has roots \(\alpha\) and \(\beta\). Find a quadratic equation with roots \(\frac { \alpha } { \beta }\) and \(\frac { \beta } { \alpha }\).
Obtain expression for \(u = \frac{\alpha}{\beta}\) in terms of \(k\) and \(\alpha\) or \(k\) and \(\beta\)
M1
A1
Obtain a correct expression
A1
Rearrange to get \(\alpha\) in terms of \(u\)
M1
Substitute into given equation
A1
Obtain correct answer
$\alpha + \beta = -k$ | B1 | State or use correct value
$\alpha\beta = 2k$ | B1 | State or use correct value
| M1 | Attempt to express sum of new roots in terms of $\alpha + \beta$, $\alpha\beta$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$ | A1 | Obtain correct expression
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{1}{2}(k-4)$ | A1 | Obtain correct answer a.e.f.
$\alpha'\beta' = 1$ | B1 | Correct product of new roots seen
$x^2 - \frac{1}{2}(k-4)x + 1 = 0$ | B1ft | Obtain correct answer, must be an eqn.
| **7 marks**
**Alternative for last 5 marks:**
Obtain expression for $u = \frac{\alpha}{\beta}$ in terms of $k$ and $\alpha$ or $k$ and $\beta$ | M1 |
| A1 | Obtain a correct expression
| A1 | Rearrange to get $\alpha$ in terms of $u$
| M1 | Substitute into given equation
| A1 | Obtain correct answer
8 The quadratic equation $x ^ { 2 } + k x + 2 k = 0$, where $k$ is a non-zero constant, has roots $\alpha$ and $\beta$. Find a quadratic equation with roots $\frac { \alpha } { \beta }$ and $\frac { \beta } { \alpha }$.
\hfill \mbox{\textit{OCR FP1 2008 Q8 [7]}}