Welcome to the World of Roots: Deep Dive into Quadratic Equations (FP1)
Hey there, brilliant mathematician! This chapter, Roots of Quadratic Equations, is absolutely foundational for Further Pure Mathematics. Don't worry if algebra sometimes feels like untangling spaghetti—we are going to break down these concepts step-by-step.
What are we doing here? Instead of solving quadratics using the complicated quadratic formula every time, we learn the hidden connections between the coefficients (the numbers \(a, b, c\)) and the roots (\(\alpha\) and \(\beta\)). This allows us to solve advanced problems much faster, especially when dealing with complex numbers.
Let's get started!
1. Understanding the Standard Quadratic Form
Every quadratic equation can be written in the general form:
\[ ax^2 + bx + c = 0 \]
where \(a\), \(b\), and \(c\) are constants, and \(a \neq 0\).
The roots of the equation, often denoted by the Greek letters \(\alpha\) (alpha) and \(\beta\) (beta), are the values of \(x\) that make the equation true. These are the points where the graph crosses the x-axis.
Prerequisite Review: Normalised Form
To find the fundamental relationships, it is often easiest to first divide the entire equation by \(a\) so that the coefficient of \(x^2\) is 1. This is called the normalised form:
\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \]
We will use this normalised form to establish our key relationships!
Quick Review Box:
The roots \(\alpha\) and \(\beta\) are the solutions to \(ax^2 + bx + c = 0\).
2. The Fundamental Relationships between Roots and Coefficients
This is the most crucial part of the chapter. Because we know that a quadratic equation can also be written in factored form using its roots:
\[ (x - \alpha)(x - \beta) = 0 \]
Expanding this gives us:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]
Now, let's compare this expanded form to our normalised form: \(x^2 + \frac{b}{a}x + \frac{c}{a} = 0\). By matching the coefficients, we get the two magic formulas:
2.1. The Sum of the Roots (\(\alpha + \beta\))
The coefficient of \(x\) in the expanded form is \( -(\alpha + \beta) \). The coefficient of \(x\) in the normalised form is \( \frac{b}{a} \). Therefore:
Formula for the Sum (S):
\[ \alpha + \beta = -\frac{b}{a} \]
2.2. The Product of the Roots (\(\alpha \beta\))
The constant term in the expanded form is \( \alpha\beta \). The constant term in the normalised form is \( \frac{c}{a} \). Therefore:
Formula for the Product (P):
\[ \alpha \beta = \frac{c}{a} \]
Memory Aid – The SOP Rule:
Think of the letters SOP (Sum, Opposite, Product).
- Sum is the opposite sign of \(\frac{b}{a}\).
- Product is exactly \(\frac{c}{a}\).
Example: Find the sum and product of the roots of \(3x^2 - 6x + 5 = 0\).
Here, \(a=3\), \(b=-6\), \(c=5\).
Sum (\(S\)): \(-\frac{b}{a} = -\frac{-6}{3} = \frac{6}{3} = 2\)
Product (\(P\)): \(\frac{c}{a} = \frac{5}{3}\)
Key Takeaway: You never need to actually solve the quadratic to find the sum or product of its roots!
3. Calculating Complex Expressions Involving Roots
In FP1, you will often be asked to find expressions like \(\alpha^2 + \beta^2\) or \(\frac{1}{\alpha} + \frac{1}{\beta}\) using only the values of \(S\) and \(P\). You must rewrite these complex expressions entirely in terms of \((\alpha + \beta)\) and \((\alpha\beta)\).
3.1. Fundamental Identity: \(\alpha^2 + \beta^2\)
Don't fall into the trap of thinking \(\alpha^2 + \beta^2 = (\alpha + \beta)^2\)! Remember the binomial expansion:
\[ (\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 \]
To isolate \(\alpha^2 + \beta^2\), we simply subtract \(2\alpha\beta\):
The Golden Identity:
\[ \alpha^2 + \beta^2 \equiv (\alpha + \beta)^2 - 2\alpha\beta \]
(This identity is essential—memorise it!)
3.2. Expressions Involving Fractions
When dealing with fractions, always combine them into a single fraction using a common denominator. This magically brings the sum and product into play.
Example: Find \(\frac{1}{\alpha} + \frac{1}{\beta}\).
Step 1: Find common denominator (\(\alpha\beta\)).
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha\beta} + \frac{\alpha}{\alpha\beta} \]
Step 2: Combine the numerators.
\[ \frac{\beta + \alpha}{\alpha\beta} = \frac{\text{Sum}}{\text{Product}} \]
3.3. Higher Power Expressions: \(\alpha^3 + \beta^3\)
This is slightly trickier, but follows the same principle: use the factored forms and substitute \(S\) and \(P\).
Recall the sum of cubes factorization: \(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\)
Substitute the Golden Identity for \(\alpha^2 + \beta^2\):
\[ \alpha^3 + \beta^3 = (\alpha + \beta) [ ((\alpha + \beta)^2 - 2\alpha\beta) - \alpha\beta ] \]
Formula for Cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta) [ (\alpha + \beta)^2 - 3\alpha\beta ] \]
!! Common Mistake to Avoid !!
When dealing with \(( \alpha - \beta )^2\), students often forget the relationship to \(S\) and \(P\). Use this trick:
\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \]
Why does this work? Because \((\alpha^2 - 2\alpha\beta + \beta^2) = (\alpha^2 + 2\alpha\beta + \beta^2) - 4\alpha\beta\).
Key Takeaway: Every expression involving \(\alpha\) and \(\beta\) must be re-written exclusively in terms of \((\alpha + \beta)\) and \((\alpha\beta)\).
4. Forming a New Quadratic Equation from Transformed Roots
The ultimate goal in this section is usually to construct a brand new quadratic equation whose roots are related to the original roots \(\alpha\) and \(\beta\).
The Universal Rule for Building a Quadratic:
If \(S'\) is the sum of the new roots and \(P'\) is the product of the new roots, the new equation is always:
\[ x^2 - S'x + P' = 0 \]
(Notice the negative sign before the sum, \(S'\)!)
4.1. Step-by-Step Process for Forming a New Equation
Suppose the original equation is \(x^2 - 5x + 6 = 0\), with roots \(\alpha\) and \(\beta\). We want a new equation whose roots are \(\alpha' = (\alpha + 2)\) and \(\beta' = (\beta + 2)\).
Step 1: Find the original Sum (S) and Product (P).
From \(x^2 - 5x + 6 = 0\):
\(a=1, b=-5, c=6\).
\(S = \alpha + \beta = -\frac{-5}{1} = 5\)
\(P = \alpha\beta = \frac{6}{1} = 6\)
Step 2: Calculate the new Sum (S').
The new roots are \(\alpha' = \alpha + 2\) and \(\beta' = \beta + 2\).
\[ S' = \alpha' + \beta' = (\alpha + 2) + (\beta + 2) \]
\[ S' = (\alpha + \beta) + 4 = S + 4 = 5 + 4 = 9 \]
Step 3: Calculate the new Product (P').
\[ P' = \alpha' \beta' = (\alpha + 2)(\beta + 2) \]
Expand carefully:
\[ P' = \alpha\beta + 2\alpha + 2\beta + 4 \]
Factor out the 2:
\[ P' = \alpha\beta + 2(\alpha + \beta) + 4 \]
Substitute \(S=5\) and \(P=6\):
\[ P' = 6 + 2(5) + 4 = 6 + 10 + 4 = 20 \]
Step 4: Form the New Equation.
Use the rule \(x^2 - S'x + P' = 0\):
\[ x^2 - 9x + 20 = 0 \]
Did you know?
The technique of transforming roots is not just an academic exercise! In advanced numerical analysis and engineering, transformations are used to shift or scale the roots of complex polynomial equations to make them easier to analyze or solve iteratively on a computer.
5. Dealing with Complex Roots
Remember that the roots of a quadratic equation can be complex numbers (in the form \(z = x + iy\)).
Key FP1 Fact: If the coefficients \(a, b, c\) are real numbers, and one root of the quadratic equation is a complex number, say \(\alpha = p + iq\), then the other root \(\beta\) must be its complex conjugate, \(\beta = p - iq\).
This is incredibly helpful because the sum and product of complex conjugates are always real numbers:
- Sum (\(\alpha + \beta\)): \((p + iq) + (p - iq) = 2p\) (Real)
- Product (\(\alpha\beta\)): \((p + iq)(p - iq) = p^2 - (iq)^2 = p^2 + q^2\) (Real)
Example: If the equation \(x^2 + bx + c = 0\) has a root \(4 + 3i\), what are \(b\) and \(c\)?
1. The roots are \(\alpha = 4 + 3i\) and \(\beta = 4 - 3i\).
2. Sum \(S = (4 + 3i) + (4 - 3i) = 8\).
3. Product \(P = (4 + 3i)(4 - 3i) = 4^2 + 3^2 = 16 + 9 = 25\).
4. Since \(S = -\frac{b}{a}\) and \(a=1\), then \(b = -S = -8\).
5. Since \(P = \frac{c}{a}\) and \(a=1\), then \(c = P = 25\).
The equation is \(x^2 - 8x + 25 = 0\).
Encouragement: Complex numbers make the algebra look intimidating, but using the sum/product relationships actually simplifies the process dramatically, especially when conjugates are involved!
Chapter Summary and Final Tips
You have mastered the core skill of relating coefficients to the roots of a quadratic equation. Keep these four formulas close at hand:
Quick Review Box: The Essentials
If \(ax^2 + bx + c = 0\) has roots \(\alpha\) and \(\beta\):
1. Sum: \(\alpha + \beta = -\frac{b}{a}\)
2. Product: \(\alpha\beta = \frac{c}{a}\)
3. The Golden Identity: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
4. New Equation Form: \(x^2 - (\text{New Sum})x + (\text{New Product}) = 0\)
Good luck with your practice! By focusing on finding the sum and product first, all other problems in this chapter become simple substitution tasks.