Edexcel IAL - Further Pure Mathematics 1- 2.2 Expressions in Roots- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 1- 2.2 Expressions in Roots -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 1- 2.2 Expressions in Roots -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 2.2 Expressions in Roots
Manipulation of Expressions Involving the Sum and Product of Roots
Let a quadratic equation be:
\( ax^2 + bx + c = 0 \)
with roots \( \alpha \) and \( \beta \).
From the coefficient–root relationships:
\( \alpha + \beta = -\dfrac{b}{a} \), \( \alpha\beta = \dfrac{c}{a} \)
Many expressions involving the roots can be written entirely in terms of these two quantities without finding the roots themselves.
Why This Is Useful
In many exam questions, finding the actual roots is unnecessary and inefficient. Instead, expressions such as \( \alpha^2 + \beta^2 \), \( \alpha^3 + \beta^3 \), and \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} \) can be evaluated using identities.
Important Identities
| Expression | In Terms of \( \alpha + \beta \) and \( \alpha\beta \) |
| \( \alpha^2 + \beta^2 \) | \( (\alpha + \beta)^2 – 2\alpha\beta \) |
| \( \alpha^3 + \beta^3 \) | \( (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta) \) |
| \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} \) | \( \dfrac{\alpha + \beta}{\alpha\beta} \) |
Identity for \( \alpha^3 + \beta^3 \)
This identity comes from algebraic expansion:
\( (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta) \)
Rearranging gives:
\( \alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta) \)
Example
The roots of \( x^2 – 5x + 6 = 0 \) are \( \alpha \) and \( \beta \). Find \( \alpha^2 + \beta^2 \).
▶️ Answer / Explanation
\( \alpha + \beta = 5 \), \( \alpha\beta = 6 \)
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \)
\( = 25 – 12 = 13 \)
Example
The roots of \( 2x^2 – 3x + 1 = 0 \) are \( \alpha \) and \( \beta \). Find \( \alpha^3 + \beta^3 \).
▶️ Answer / Explanation
\( \alpha + \beta = \dfrac{3}{2} \), \( \alpha\beta = \dfrac{1}{2} \)
\( \alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta) \)
\( = \left(\dfrac{3}{2}\right)^3 – 3\left(\dfrac{1}{2}\right)\left(\dfrac{3}{2}\right) \)
\( = \dfrac{27}{8} – \dfrac{9}{4} = \dfrac{9}{8} \)
Example
If the roots of \( x^2 + 4x + 5 = 0 \) are \( \alpha \) and \( \beta \), find \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} \).
▶️ Answer / Explanation
\( \alpha + \beta = -4 \), \( \alpha\beta = 5 \)
\( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-4}{5} \)