Standard +0.8 This question requires understanding that 'two pairs of equal roots' means the quartic factors as (x-α)²(x-β)², then expanding and comparing coefficients to derive the given relationship. While the algebraic manipulation is straightforward once the structure is recognized, the conceptual leap to interpret the condition and set up the problem correctly elevates this above a routine exercise. It's a standard Further Maths question requiring insight but not extensive calculation.
1 The quartic equation \(x ^ { 4 } - p x ^ { 2 } + q x - r = 0\), where \(p , q\) and \(r\) are real constants, has two pairs of equal roots. Show that \(p ^ { 2 } + 4 r = 0\) and state the value of \(q\).
## Question 1:
| Working/Answer | Marks | Notes |
|---|---|---|
| i. $2\alpha + 2\beta = 0$ | B2 (any two correct) | All four correct for B2 |
| ii. $\alpha^2 + 4\alpha\beta + \beta^2 = -p$ | (B1 for any two correct) | |
| iii. $2\alpha^2\beta + 2\alpha\beta^2 = -q$ | | |
| iv. $\alpha^2\beta^2 = -r$ | | |
| Use of $\beta = -\alpha$ in ii, iii or iv | M1 | |
| $p = 2\alpha^2$ and $\alpha^4 = -r \Rightarrow p^2 + 4r = 0$ | M1A1 | AG |
| $q = 0$ | B1 | CAO |
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1 The quartic equation $x ^ { 4 } - p x ^ { 2 } + q x - r = 0$, where $p , q$ and $r$ are real constants, has two pairs of equal roots. Show that $p ^ { 2 } + 4 r = 0$ and state the value of $q$.
\hfill \mbox{\textit{CAIE FP1 2015 Q1 [6]}}