AS 1 Pure Mathematics: Roots of Quadratic Equations
Welcome to Further Algebra and Functions for CCEA AS 1 Pure Mathematics! In GCSE and standard AS Mathematics, you learned how to solve quadratic equations using factorisation, the quadratic formula, or completing the square. In Further Mathematics, we look at equations from a powerful new perspective: exploring the relationships between the roots and their coefficients without needing to solve the equations at all.
Don't worry if this seems abstract at first! Think of coefficients as "clues" and roots as "hidden identities". By learning a few simple algebraic tools, you will be able to unlock values, manipulate symmetric expressions, and build entirely new equations with confidence.
1. Relations Between Roots and Coefficients
Consider a general quadratic equation:
\(ax^2 + bx + c = 0 \quad (a \neq 0)\)
Let the two roots of this equation be called \(\alpha\) (alpha) and \(\beta\) (beta). If \(\alpha\) and \(\beta\) are the roots, then by the Factor Theorem, the equation can be written in factorised form as:
\(a(x - \alpha)(x - \beta) = 0\)
Expanding this expression gives:
\(a(x^2 - (\alpha + \beta)x + \alpha\beta) = 0\)
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
Now, divide the original equation \(ax^2 + bx + c = 0\) through by \(a\):
\(x^2 + \frac{b}{a}x + \frac{c}{a} = 0\)
By comparing the coefficients of both forms, we get two fundamental relations:
1. Sum of roots:
\(\alpha + \beta = -\frac{b}{a}\)
2. Product of roots:
\(\alpha\beta = \frac{c}{a}\)
Memory Tip: Remember that the Sum always has the negative sign (\(-\frac{b}{a}\)), while the Product keeps the positive sign (\(+\frac{c}{a}\)).
Important Examination Note: These two formulae are NOT provided in the CCEA GCE Mathematical Formulae and Tables booklet. You are expected to know them by heart!
Worked Example: Finding the Sum and Product
Question: For the quadratic equation \(3x^2 - 7x + 2 = 0\), write down the value of \(\alpha + \beta\) and \(\alpha\beta\).
Solution:
Identify the coefficients: \(a = 3\), \(b = -7\), \(c = 2\).
Using the relations:
\(\alpha + \beta = -\frac{b}{a} = -\frac{-7}{3} = \frac{7}{3}\)
\(\alpha\beta = \frac{c}{a} = \frac{2}{3}\)
Section Takeaway: For any quadratic \(ax^2 + bx + c = 0\), the sum of the roots is always \(-\frac{b}{a}\) and the product is always \(\frac{c}{a}\).
2. Symmetric Functions of Roots
A symmetric function of \(\alpha\) and \(\beta\) is an algebraic expression that remains completely unchanged if you swap \(\alpha\) and \(\beta\). For example, \(\alpha^2 + \beta^2\) is symmetric because swapping gives \(\beta^2 + \alpha^2\), which is identical.
In the exam, you will frequently be asked to evaluate expressions involving \(\alpha\) and \(\beta\). The key strategy is to rewrite the expression purely in terms of \((\alpha + \beta)\) and \((\alpha\beta)\).
Essential Standard Identities
1. Sum of Squares: \(\alpha^2 + \beta^2\)
Start with the expansion of \((\alpha + \beta)^2\):
\((\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2\)
Rearranging to isolate \(\alpha^2 + \beta^2\):
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
2. Sum of Cubes: \(\alpha^3 + \beta^3\)
Start with the binomial expansion of \((\alpha + \beta)^3\):
\((\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)\)
Rearranging to isolate \(\alpha^3 + \beta^3\):
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
3. Difference of Roots Squared: \((\alpha - \beta)^2\)
Expanding gives:
\((\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2 = (\alpha^2 + \beta^2) - 2\alpha\beta\)
Substitute the sum of squares identity:
\((\alpha - \beta)^2 = [(\alpha + \beta)^2 - 2\alpha\beta] - 2\alpha\beta\)
\((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
4. Sum of Reciprocals: \(\frac{1}{\alpha} + \frac{1}{\beta}\)
Find a common denominator:
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}\)
5. Sum of Reciprocal Squares: \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\)
Combine over a common denominator and apply the sum of squares identity:
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}\)
Worked Example: Evaluating Symmetric Expressions
Question: The roots of the quadratic equation \(2x^2 + 5x - 4 = 0\) are \(\alpha\) and \(\beta\). Without solving the equation, find the exact values of:
(a) \(\alpha^2 + \beta^2\)
(b) \(\alpha^3 + \beta^3\)
(c) \(\frac{1}{\alpha} + \frac{1}{\beta}\)
Solution:
First, find the basic sum and product of the roots:
\(\alpha + \beta = -\frac{b}{a} = -\frac{5}{2}\)
\(\alpha\beta = \frac{c}{a} = \frac{-4}{2} = -2\)
Part (a):
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(\alpha^2 + \beta^2 = \left(-\frac{5}{2}\right)^2 - 2(-2) = \frac{25}{4} + 4 = \frac{25}{4} + \frac{16}{4} = \frac{41}{4}\)
Part (b):
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(\alpha^3 + \beta^3 = \left(-\frac{5}{2}\right)^3 - 3(-2)\left(-\frac{5}{2}\right)\)
\(\alpha^3 + \beta^3 = -\frac{125}{8} - 15 = -\frac{125}{8} - \frac{120}{8} = -\frac{245}{8}\)
Part (c):
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-\frac{5}{2}}{-2} = \frac{5}{4}\)
Section Takeaway: Whenever you see combinations of roots like \(\alpha^2 + \beta^2\) or \(\frac{1}{\alpha} + \frac{1}{\beta}\), rewrite them in terms of \((\alpha + \beta)\) and \((\alpha\beta)\) before substituting your numerical values.
3. Forming New Quadratic Equations
A classic exam question asks you to form a new quadratic equation whose roots are related to the original roots \(\alpha\) and \(\beta\) (for example, roots given by \(u\) and \(v\), such as \(2\alpha\) and \(2\beta\), or \(\alpha^2\) and \(\beta^2\)).
The 4-Step Method for Forming Equations
Step 1: Calculate the New Sum (\(S\))
Find the sum of the new roots: \(S = u + v\). Express this in terms of \((\alpha + \beta)\) and \((\alpha\beta)\), then evaluate.
Step 2: Calculate the New Product (\(P\))
Find the product of the new roots: \(P = uv\). Express this in terms of \((\alpha + \beta)\) and \((\alpha\beta)\), then evaluate.
Step 3: Substitute into the Standard Quadratic Structure
Substitute \(S\) and \(P\) into the general equation structure:
\(x^2 - Sx + P = 0 \quad \text{or} \quad x^2 - (\text{Sum})x + (\text{Product}) = 0\)
Step 4: Clear Denominators (if requested)
If the question specifies integer coefficients (\(a, b, c \in \mathbb{Z}\)), multiply the entire equation by the lowest common denominator to eliminate fractions.
Worked Example: Constructing a New Equation
Question: The roots of the quadratic equation \(x^2 - 4x + 6 = 0\) are \(\alpha\) and \(\beta\). Find a quadratic equation with integer coefficients whose roots are \(\frac{1}{\alpha^2}\) and \(\frac{1}{\beta^2}\).
Solution:
From the original equation \(x^2 - 4x + 6 = 0\):
\(\alpha + \beta = -\frac{-4}{1} = 4\)
\(\alpha\beta = \frac{6}{1} = 6\)
Step 1: Find the New Sum (\(S\))
\(S = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}\)
\(S = \frac{(4)^2 - 2(6)}{(6)^2} = \frac{16 - 12}{36} = \frac{4}{36} = \frac{1}{9}\)
Step 2: Find the New Product (\(P\))
\(P = \left(\frac{1}{\alpha^2}\right)\left(\frac{1}{\beta^2}\right) = \frac{1}{(\alpha\beta)^2}\)
\(P = \frac{1}{(6)^2} = \frac{1}{36}\)
Step 3: Write the equation
\(x^2 - Sx + P = 0\)
\(x^2 - \frac{1}{9}x + \frac{1}{36} = 0\)
Step 4: Clear denominators for integer coefficients
Multiply every term by \(36\):
\(36x^2 - 4x + 1 = 0\)
Section Takeaway: Building new quadratics always follows the template \(x^2 - Sx + P = 0\). Always remember to include the \("= 0"\) at the end!
4. Pitfalls & Examiner-Reported Errors
To secure full marks on Unit AS 1, watch out for these very common mistakes highlighted in examiner reports:
1. Missing Negative Sign in the Sum:
Writing \(\alpha + \beta = \frac{b}{a}\) instead of \(-\frac{b}{a}\). If \(b\) is already negative, remember that \(-\frac{-b}{a}\) becomes positive.
2. Wrong Sign in the New Equation:
Writing \(x^2 + Sx + P = 0\) instead of \(x^2 - Sx + P = 0\). The minus sign before the sum term is essential.
3. Forgetting the \("= 0"\):
If asked to "find an equation", writing just an expression like \(36x^2 - 4x + 1\) will lose the final accuracy mark. You must write \(36x^2 - 4x + 1 = 0\).
4. Expansion Errors with Cubes:
Assuming that \(\alpha^3 + \beta^3 = (\alpha + \beta)^3\). Remember the extra adjustment term: \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\).
5. Leaving Fractions in Integer Form:
If the question specifies coefficients in \(\mathbb{Z}\) (integers), leaving \(x^2 - \frac{1}{9}x + \frac{1}{36} = 0\) is incomplete. Always multiply through by the common denominator.
6. Rounding Accuracy:
Unless exact values (fractions/surds) are requested, any calculated decimal values must be given to 3 significant figures.
5. Chapter Summary Checklist
Before sitting your AS 1 examination, make sure you can confidently:
1. State from memory that \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\).
2. Expand and rewrite \(\alpha^2 + \beta^2\), \(\alpha^3 + \beta^3\), and \((\alpha - \beta)^2\) in terms of sum and product.
3. Combine algebraic fractions involving \(\alpha\) and \(\beta\) over common denominators.
4. Calculate the new sum \(S\) and new product \(P\) to form \(x^2 - Sx + P = 0\).
5. Check that your final answer is an equation with \("= 0"\) and has integer coefficients if required.