8. The roots of the cubic equation \(x ^ { 3 } + 5 x ^ { 2 } + 2 x + 8 = 0\) are denoted by \(\alpha , \beta , \gamma\).
Determine the cubic equation whose roots are \(\frac { \alpha } { \beta \gamma } , \frac { \beta } { \gamma \alpha } , \frac { \gamma } { \alpha \beta }\).
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\(\alpha + \beta + \gamma = -5\), \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\), \(\alpha\beta\gamma = -8\) B1
any two correct equations
New equation:
Answer Marks
Guidance
Sum of roots: \(\frac{\alpha}{\beta\gamma} + \frac{\beta}{\gamma\alpha} + \frac{\gamma}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta\gamma}\) M1
Common denominator
\(= \frac{(\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)}{\alpha\beta\gamma} = \frac{(-5)^2 - (2 \times 2)}{-8} = \frac{21}{-8}\) A1
Sum of pairs:
Answer Marks
Guidance
\(\frac{1}{\gamma^2} + \frac{1}{\beta^2} + \frac{1}{\alpha^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \alpha^2\gamma^2}{\alpha^2\beta^2\gamma^2}\) M1
Common denominator
\(= \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}\) A1
Fully factorised
\(= \frac{2^2 - (2 \times -8 \times -5)}{(-8)^2} = \frac{-76}{64} = \left(\frac{-19}{16}\right)\) A1
Product:
Answer Marks
Guidance
\(\frac{1}{\alpha\beta\gamma} = \frac{-1}{8}\) B1
\(\therefore \frac{-b}{a} = \frac{21}{8}\), \(\frac{c}{a} = \frac{-76}{64}\), \(\frac{-d}{a} = \frac{-1}{8}\) B1
FT previous values, two correct expression
If \(a = 1\), \(b = \frac{21}{8}\), \(c = \frac{-76}{64}\), \(d = \frac{1}{8}\)
Answer Marks
Guidance
New equation: \(x^3 + \frac{21}{8}x^2 - \frac{76}{64}x + \frac{1}{8} = 0\) B1
oe; FT B1 above; eg If \(a = 16\), \(b = 42\), \(c = -19\), \(d = 2\) New equation: \(16x^3 + 42x^2 - 19x + 2 = 0\)
Total: [9] Question 9a
Answer Marks
\(u + iv = 1 - (x + iy)^2\), \(u + iv = 1 - x^2 + y^2 - 2ixy\) M1
Comparing coefficients: Imaginary parts: \(v = -2xy\), Real parts: \(u = 1 - x^2 + y^2\) A1
[4] Question 9b
Answer Marks
Guidance
Substituting \(y = 4x\): \(v = -2x \times 4x = -8x^2\), \(u = 1 - x^2 + 16x^2 = 1 + 15x^2\) M1
FT (a)
A1 A1 for both \(u\) and \(v\)
Eliminating \(x\), the equation of the locus \(Q\) is \(u = 1 + 15\left(\frac{v}{-8}\right)\) oe M1
A1 cao \((8u + 15v = 8)\)
[4] Question 9c
Answer Marks
Guidance
Point \(P(2,5) \to Q(22, -20)\) B1
FT (a)
Equation of the locus of \(Q\) is \(8u + 15v = 8\)
Answer Marks
Guidance
\(D = \frac{ (8 \times 22) + (15 \times -20) - 8
}{\sqrt{64 + 225}}\)
A1 oe FT their \(Q\) (not \(P\)) & straight line from (b)
\(D = \frac{132}{17}\) or \(7.7647 \ldots\) A1
cao
[4] Total: [12]
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$\alpha + \beta + \gamma = -5$, $\alpha\beta + \beta\gamma + \gamma\alpha = 2$, $\alpha\beta\gamma = -8$ | B1 | any two correct equations
**New equation:**
Sum of roots: $\frac{\alpha}{\beta\gamma} + \frac{\beta}{\gamma\alpha} + \frac{\gamma}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta\gamma}$ | M1 | Common denominator
$= \frac{(\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)}{\alpha\beta\gamma} = \frac{(-5)^2 - (2 \times 2)}{-8} = \frac{21}{-8}$ | A1 |
**Sum of pairs:**
$\frac{1}{\gamma^2} + \frac{1}{\beta^2} + \frac{1}{\alpha^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \alpha^2\gamma^2}{\alpha^2\beta^2\gamma^2}$ | M1 | Common denominator
$= \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$ | A1 | Fully factorised
$= \frac{2^2 - (2 \times -8 \times -5)}{(-8)^2} = \frac{-76}{64} = \left(\frac{-19}{16}\right)$ | A1 |
**Product:**
$\frac{1}{\alpha\beta\gamma} = \frac{-1}{8}$ | B1 |
$\therefore \frac{-b}{a} = \frac{21}{8}$, $\frac{c}{a} = \frac{-76}{64}$, $\frac{-d}{a} = \frac{-1}{8}$ | B1 | FT previous values, two correct expression
If $a = 1$, $b = \frac{21}{8}$, $c = \frac{-76}{64}$, $d = \frac{1}{8}$
New equation: $x^3 + \frac{21}{8}x^2 - \frac{76}{64}x + \frac{1}{8} = 0$ | B1 | oe; FT B1 above; eg If $a = 16$, $b = 42$, $c = -19$, $d = 2$ New equation: $16x^3 + 42x^2 - 19x + 2 = 0$
**Total: [9]**
# Question 9a
$u + iv = 1 - (x + iy)^2$, $u + iv = 1 - x^2 + y^2 - 2ixy$ | M1 |
Comparing coefficients: Imaginary parts: $v = -2xy$, Real parts: $u = 1 - x^2 + y^2$ | A1 |
**[4]**
# Question 9b
Substituting $y = 4x$: $v = -2x \times 4x = -8x^2$, $u = 1 - x^2 + 16x^2 = 1 + 15x^2$ | M1 | FT (a)
| A1 | A1 for both $u$ and $v$
Eliminating $x$, the equation of the locus $Q$ is $u = 1 + 15\left(\frac{v}{-8}\right)$ oe | M1 |
| A1 | cao $(8u + 15v = 8)$
**[4]**
# Question 9c
Point $P(2,5) \to Q(22, -20)$ | B1 | FT (a)
Equation of the locus of $Q$ is $8u + 15v = 8$
$D = \frac{|(8 \times 22) + (15 \times -20) - 8|}{\sqrt{64 + 225}}$ | M1 |
| A1 | oe FT their $Q$ (not $P$) & straight line from (b)
$D = \frac{132}{17}$ or $7.7647 \ldots$ | A1 | cao
**[4]**
**Total: [12]**
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8. The roots of the cubic equation $x ^ { 3 } + 5 x ^ { 2 } + 2 x + 8 = 0$ are denoted by $\alpha , \beta , \gamma$.
Determine the cubic equation whose roots are $\frac { \alpha } { \beta \gamma } , \frac { \beta } { \gamma \alpha } , \frac { \gamma } { \alpha \beta }$.\\
Give your answer in the form $a x ^ { 3 } + b x ^ { 2 } + c x + d = 0$, where $a , b , c , d$ are constants to be determined.\\
\hfill \mbox{\textit{WJEC Further Unit 1 2023 Q8 [9]}}