Edexcel IAL - Further Pure Mathematics 1- 2.3 Forming Quadratic Equations with New Roots- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 1- 2.3 Forming Quadratic Equations with New Roots -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 1- 2.3 Forming Quadratic Equations with New Roots -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 2.3 Forming Quadratic Equations with New Roots
Forming Quadratic Equations with New Roots
Let the quadratic equation
\( ax^2 + bx + c = 0 \)
have roots \( \alpha \) and \( \beta \).
Then:
\( \alpha + \beta = -\dfrac{b}{a} \), \( \alpha\beta = \dfrac{c}{a} \)
Using only these two values, we can form a new quadratic equation whose roots are functions of \( \alpha \) and \( \beta \), without finding the actual roots.
General Method
If the new roots are \( p \) and \( q \), then the required quadratic equation is:
\( x^2 – (p + q)x + pq = 0 \)
So the task reduces to finding the sum and product of the new roots in terms of \( \alpha + \beta \) and \( \alpha\beta \).
Common New Roots
| New roots | Sum and Product |
| \( \dfrac{1}{\alpha},\ \dfrac{1}{\beta} \) | Sum \( = \dfrac{\alpha + \beta}{\alpha\beta} \), Product \( = \dfrac{1}{\alpha\beta} \) |
| \( \alpha^2,\ \beta^2 \) | Sum \( = (\alpha + \beta)^2 – 2\alpha\beta \), Product \( = (\alpha\beta)^2 \) |
| \( \alpha^3,\ \beta^3 \) | Sum \( = (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta) \), Product \( = (\alpha\beta)^3 \) |
Example
The roots of \( x^2 – 5x + 6 = 0 \) are \( \alpha \) and \( \beta \). Form the quadratic equation whose roots are \( \dfrac{1}{\alpha} \) and \( \dfrac{1}{\beta} \).
▶️ Answer / Explanation
\( \alpha + \beta = 5 \), \( \alpha\beta = 6 \)
New sum:
\( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{6} \)
New product:
\( \dfrac{1}{\alpha\beta} = \dfrac{1}{6} \)
Required equation:
\( x^2 – \dfrac{5}{6}x + \dfrac{1}{6} = 0 \)
Multiply by 6:
\( 6x^2 – 5x + 1 = 0 \)
Example
The roots of \( 2x^2 – 3x + 1 = 0 \) are \( \alpha \) and \( \beta \). Form the quadratic equation whose roots are \( \alpha^2 \) and \( \beta^2 \).
▶️ Answer / Explanation
\( \alpha + \beta = \dfrac{3}{2} \), \( \alpha\beta = \dfrac{1}{2} \)
New sum:
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = \dfrac{9}{4} – 1 = \dfrac{5}{4} \)
New product:
\( \alpha^2\beta^2 = (\alpha\beta)^2 = \dfrac{1}{4} \)
Required equation:
\( x^2 – \dfrac{5}{4}x + \dfrac{1}{4} = 0 \)
Multiply by 4:
\( 4x^2 – 5x + 1 = 0 \)
Example
The roots of \( x^2 – 4x + 3 = 0 \) are \( \alpha \) and \( \beta \). Form the quadratic equation whose roots are \( \alpha + \dfrac{2}{\beta} \) and \( \beta + \dfrac{2}{\alpha} \).
▶️ Answer / Explanation
\( \alpha + \beta = 4 \), \( \alpha\beta = 3 \)
New sum:
\( \alpha + \dfrac{2}{\beta} + \beta + \dfrac{2}{\alpha} = (\alpha + \beta) + 2\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right) \)
\( = 4 + 2\left(\dfrac{4}{3}\right) = \dfrac{20}{3} \)
New product:
\( (\alpha + \dfrac{2}{\beta})(\beta + \dfrac{2}{\alpha}) = \alpha\beta + 2 + 2 + \dfrac{4}{\alpha\beta} \)
\( = 3 + 4 + \dfrac{4}{3} = \dfrac{25}{3} \)
Required equation:
\( x^2 – \dfrac{20}{3}x + \dfrac{25}{3} = 0 \)
Multiply by 3:
\( 3x^2 – 20x + 25 = 0 \)