14 exam-style questions on relations between roots, transformations and symmetric functions. Tap Show answer for full working.
The cubic equation \(x^3 - 6x^2 + 11x - 6 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\).
Find \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\).
For \(x^3 + ax^2 + bx + c = 0\): sum of roots \(= -a\), sum of products two at a time \(= b\), product \(= -c\).
Here \(x^3 - 6x^2 + 11x - 6 = 0\):
\(\alpha+\beta+\gamma = 6\), \(\alpha\beta+\beta\gamma+\gamma\alpha = 11\), \(\alpha\beta\gamma = 6\).
The equation \(2x^3 - 5x^2 + x + 2 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\).
Find the values of \(\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}\) and \(\alpha^2+\beta^2+\gamma^2\).
Divide by 2: \(x^3 - \dfrac{5}{2}x^2 + \dfrac{1}{2}x + 1 = 0\).
\(\alpha+\beta+\gamma = \dfrac{5}{2}\), \(\alpha\beta+\beta\gamma+\gamma\alpha = \dfrac{1}{2}\), \(\alpha\beta\gamma = -1\).
\(\sum\dfrac{1}{\alpha} = \dfrac{\sum\beta\gamma}{\alpha\beta\gamma} = \dfrac{1/2}{-1} = -\dfrac{1}{2}\)
\(\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = \dfrac{25}{4} - 1 = \dfrac{21}{4}\)
The roots of \(x^3 + px^2 + qx + r = 0\) are \(\alpha\), \(\beta\), \(\gamma\).
Write down expressions for \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\) in terms of \(p\), \(q\), \(r\).
\(\alpha+\beta+\gamma = -p\)
\(\alpha\beta+\beta\gamma+\gamma\alpha = q\)
\(\alpha\beta\gamma = -r\)
The equation \(x^3 - 4x^2 + x + 6 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\).
Find the cubic equation with roots \(\alpha+1\), \(\beta+1\), \(\gamma+1\).
Let \(y = x+1\), so \(x = y-1\).
\((y-1)^3 - 4(y-1)^2 + (y-1) + 6 = 0\)
\(y^3 - 3y^2 + 3y - 1 - 4(y^2-2y+1) + y - 1 + 6 = 0\)
\(y^3 - 3y^2 + 3y - 1 - 4y^2 + 8y - 4 + y + 5 = 0\)
\(y^3 - 7y^2 + 12y = 0\)
\(y(y^2 - 7y + 12) = 0\) or \(y^3 - 7y^2 + 12y = 0\).
The roots of \(x^3 + 3x^2 - 4x - 12 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).
Find the equation with roots \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\), \(\dfrac{1}{\gamma}\).
Let \(y = 1/x\), so \(x = 1/y\).
\(\dfrac{1}{y^3} + 3\dfrac{1}{y^2} - 4\dfrac{1}{y} - 12 = 0\). Multiply by \(y^3\):
\(1 + 3y - 4y^2 - 12y^3 = 0\)
\(12y^3 + 4y^2 - 3y - 1 = 0\).
Given that \(\alpha\), \(\beta\), \(\gamma\) are roots of \(x^3 - 5x^2 + 2x + 8 = 0\), find the value of \(\alpha^3 + \beta^3 + \gamma^3\).
Each root satisfies \(\alpha^3 = 5\alpha^2 - 2\alpha - 8\), so \(\sum\alpha^3 = 5\sum\alpha^2 - 2\sum\alpha - 8\cdot 3\).
\(\sum\alpha = 5\), \(\sum\alpha\beta = 2\), \(\alpha\beta\gamma = -8\).
\(\sum\alpha^2 = 25 - 4 = 21\).
\(\sum\alpha^3 = 5\cdot21 - 2\cdot5 - 24 = 105 - 10 - 24 = 71\).
The cubic \(x^3 + ax^2 + bx + c = 0\) has roots \(2\), \(-1\) and \(k\). Find \(a\), \(b\), \(c\) and \(k\).
\((x-2)(x+1)(x-k) = 0\)
Sum of roots: \(2-1+k = -a \Rightarrow 1+k = -a\)
We need more: expand \((x-2)(x+1)(x-k) = (x^2 - x - 2)(x-k) = x^3 - (k+1)x^2 + (k-2)x + 2k\)
Without further info we cannot uniquely find \(k\). If monic with three roots only: any \(k\) works with matching coefficients.
Assuming the question implies find in terms of relations — alternatively if product etc. given elsewhere. Typical: if also sum of products given. For completeness: if only roots 2, −1, \(k\), then \(a = -(1+k)\), \(b = k-2\), \(c = -2k\).
The equation \(x^3 - 6x^2 + 11x - 6 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\).
Find the equation whose roots are \(\alpha^2\), \(\beta^2\), \(\gamma^2\).
Roots of original: 1, 2, 3 (by inspection). Then squares: 1, 4, 9.
Equation: \((x-1)(x-4)(x-9) = 0\)
\(x^3 - 14x^2 + 49x - 36 = 0\).
Without finding roots: let \(y = x^2\). From \(\sum\alpha = 6\), \(\sum\alpha\beta = 11\), \(\alpha\beta\gamma = 6\).
\(\sum\alpha^2 = 36-22 = 14\), \(\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2\alpha\beta\gamma\sum\alpha = 121 - 2\cdot6\cdot6 = 121-72 = 49\), \(\alpha^2\beta^2\gamma^2 = 36\).
So \(y^3 - 14y^2 + 49y - 36 = 0\).
Given that \(\alpha\), \(\beta\), \(\gamma\) are roots of \(x^3 + px + q = 0\) (no \(x^2\) term), show that \(\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma\).
Hence find \(\alpha^3 + \beta^3 + \gamma^3\) when the equation is \(x^3 - 3x + 2 = 0\).
\(\sum\alpha = 0\), \(\sum\alpha\beta = p\), \(\alpha\beta\gamma = -q\).
\(\sum\alpha^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta-\beta\gamma-\gamma\alpha)\).
Since \(\sum\alpha = 0\), \(\sum\alpha^3 = 3\alpha\beta\gamma\).
For \(x^3 - 3x + 2 = 0\): \(p = -3\), \(q = 2\), \(\alpha\beta\gamma = -2\).
\(\sum\alpha^3 = 3(-2) = -6\).
The roots of \(x^3 - 3x^2 - 4x + 12 = 0\) are \(\alpha\), \(\beta\), \(\gamma\).
Find the equation with roots \(\alpha+\beta\), \(\beta+\gamma\), \(\gamma+\alpha\).
\(\sum\alpha = 3\), so \(\alpha+\beta = 3-\gamma\), etc. Roots of new equation are \(3-\gamma\), \(3-\alpha\), \(3-\beta\).
This is the same transformation as \(y = 3 - x\) on the original roots (permuted).
\(x = 3-y\): \((3-y)^3 - 3(3-y)^2 - 4(3-y) + 12 = 0\)
Expand: \(27 - 27y + 9y^2 - y^3 - 3(9 - 6y + y^2) - 12 + 4y + 12 = 0\)
\(27 - 27y + 9y^2 - y^3 - 27 + 18y - 3y^2 - 12 + 4y + 12 = 0\)
\(-y^3 + 6y^2 - 5y = 0\)
\(y^3 - 6y^2 + 5y = 0\) or \(y(y^2 - 6y + 5) = 0\).
The quadratic equation \( 2x^2 - 5x + 3 = 0 \) has roots \( \alpha \) and \( \beta \). Find the value of \( \alpha^2 + \beta^2 \) and of \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} \).
\( \alpha + \beta = 5/2 \), \( \alpha\beta = 3/2 \).
\( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (25/4) - 3 = 13/4 \).
\( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{5/2}{3/2} = 5/3 \).
The cubic equation \( x^3 + px^2 + qx + r = 0 \) has roots \( \alpha, \beta, \gamma \). Express \( \alpha^2 + \beta^2 + \gamma^2 \) and \( \alpha^3 + \beta^3 + \gamma^3 \) in terms of \( p, q, r \).
\( \alpha+\beta+\gamma = -p \), \( \alpha\beta+\beta\gamma+\gamma\alpha = q \), \( \alpha\beta\gamma = -r \).
\( \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = p^2 - 2q \).
From the equation, \( \alpha^3 = -p\alpha^2 - q\alpha - r \), so summing: \( \alpha^3+\beta^3+\gamma^3 = -p(\alpha^2+\beta^2+\gamma^2) - q(\alpha+\beta+\gamma) - 3r \).
\( = -p(p^2-2q) - q(-p) - 3r = -p^3 + 2pq + pq - 3r = -p^3 + 3pq - 3r \).
The equation \( x^3 - 4x^2 + x + 6 = 0 \) has roots \( \alpha, \beta, \gamma \). Find the equation with roots \( \alpha+1 \), \( \beta+1 \), \( \gamma+1 \).
Let \( y = x + 1 \), so \( x = y - 1 \).
\( (y-1)^3 - 4(y-1)^2 + (y-1) + 6 = 0 \)
\( y^3 - 3y^2 + 3y - 1 - 4(y^2 - 2y + 1) + y - 1 + 6 = 0 \)
\( y^3 - 3y^2 + 3y - 1 - 4y^2 + 8y - 4 + y - 1 + 6 = 0 \)
\( y^3 - 7y^2 + 12y = 0 \)
\( y(y^2 - 7y + 12) = 0 \) or the monic cubic \( y^3 - 7y^2 + 12y = 0 \).
The cubic \( x^3 + ax^2 + bx + c = 0 \) has roots \( \alpha, \beta, \gamma \) such that \( \alpha + \beta = \gamma \). Show that \( a^3 - 4ab + 8c = 0 \) and find a condition on \( a,b,c \) for the roots to be real.
\( \alpha+\beta+\gamma = -a \) and \( \alpha+\beta = \gamma \) ⇒ \( 2\gamma = -a \) ⇒ \( \gamma = -a/2 \), and \( \alpha+\beta = -a/2 \).
Also \( \alpha\beta + \gamma(\alpha+\beta) = b \) ⇒ \( \alpha\beta + (-a/2)(-a/2) = b \) ⇒ \( \alpha\beta + a^2/4 = b \) ⇒ \( \alpha\beta = b - a^2/4 \).
Product \( \alpha\beta\gamma = -c \) ⇒ \( (b - a^2/4)(-a/2) = -c \) ⇒ \( -(a/2)(b - a^2/4) = -c \) ⇒ \( (a/2)(b - a^2/4) = c \).
Multiply by 8: \( 4a(b - a^2/4) = 8c \) ⇒ \( 4ab - a^3 = 8c \) ⇒ \( a^3 - 4ab + 8c = 0 \).
For real roots: \( \alpha, \beta \) are roots of \( t^2 - (\alpha+\beta)t + \alpha\beta = 0 \) i.e. \( t^2 + (a/2)t + (b - a^2/4) = 0 \). Discriminant \( \geq 0 \): \( a^2/4 - 4(b - a^2/4) \geq 0 \) ⇒ \( a^2/4 - 4b + a^2 \geq 0 \) ⇒ \( (5/4)a^2 - 4b \geq 0 \) ⇒ \( 5a^2 \geq 16b \).