Introduction to Forming Equations

In your previous studies, you usually started with a quadratic equation and worked hard to find its roots (the values of \(x\)). In this chapter, we are going to do the exact opposite! We will start with the roots and "work backward" to build the original quadratic equation.

Think of it like being an architect: instead of exploring a finished building, you are given the bricks (the roots) and asked to design the house (the equation) they belong to.

The "Master Formula"

Every quadratic equation can be written in a specific way if we know its roots. Let’s say the two roots are \(\alpha\) (alpha) and \(\beta\) (beta). Any quadratic equation with these roots can be expressed as:

\(x^2 - (\alpha + \beta)x + (\alpha\beta) = 0\)

This is the most important formula in this chapter. It tells us that to form an equation, we only need to find two things:

  1. The Sum of the roots: \(\alpha + \beta\)
  2. The Product of the roots: \(\alpha\beta\)

Important Note: Always remember the minus sign before the sum of the roots! A common mistake is writing \(x^2 + (\text{sum})x\), but it must be \(x^2 - (\text{sum})x\).

Step-by-Step: Forming an Equation from Numbers

Don't worry if this seems tricky; let’s look at a simple example first. Suppose we want to form a quadratic equation with the roots \(5\) and \(-2\).

Step 1: Find the Sum.
\(5 + (-2) = 3\)

Step 2: Find the Product.
\(5 \times (-2) = -10\)

Step 3: Plug them into the formula.
\(x^2 - (3)x + (-10) = 0\)
Which simplifies to: \(x^2 - 3x - 10 = 0\)

Forming New Equations from Old Ones

In Further Pure Mathematics, the exam often asks you to form a new equation based on the roots of an existing equation. This usually involves three distinct stages.

1. The "Old" Equation

Suppose you are given \(2x^2 + 5x - 3 = 0\) with roots \(\alpha\) and \(\beta\). From the previous chapter (Functions of the roots), we know:
Sum: \(\alpha + \beta = -\frac{b}{a} = -\frac{5}{2}\)
Product: \(\alpha\beta = \frac{c}{a} = -\frac{3}{2}\)

2. The "New" Roots

The question might ask you to form a new equation with roots \(2\alpha\) and \(2\beta\). We need to find the sum and product of these new roots.

New Sum: \(2\alpha + 2\beta = 2(\alpha + \beta)\)
Using our old sum: \(2(-\frac{5}{2}) = -5\)

New Product: \((2\alpha)(2\beta) = 4\alpha\beta\)
Using our old product: \(4(-\frac{3}{2}) = -6\)

3. The "New" Equation

Now use the Master Formula with your new values:
\(x^2 - (\text{New Sum})x + (\text{New Product}) = 0\)
\(x^2 - (-5)x + (-6) = 0\)
Final Answer: \(x^2 + 5x - 6 = 0\)

Quick Review: The 3-Step Process

1. Identify \(\alpha + \beta\) and \(\alpha\beta\) from the given equation.
2. Calculate the New Sum and New Product for the roots requested.
3. Substitute into \(x^2 - (\text{Sum})x + (\text{Product}) = 0\).

Common Algebraic Identities You’ll Need

Sometimes the "new roots" are more complex, like \(\alpha^2\) and \(\beta^2\). To find their sum, you must use these identities (which you should memorize):

  • \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
  • \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)

Helpful Tips and Common Pitfalls

The "= 0" is Mandatory: An equation must have an equals sign. If you just write \(x^2 - 5x + 6\), it is an "expression," and you may lose a mark. Always include \(= 0\).

Fractional Coefficients: If your equation looks like \(x^2 - \frac{1}{2}x + \frac{3}{4} = 0\), it is often "cleaner" to multiply the whole thing by 4 to get \(4x^2 - 2x + 3 = 0\). Both are mathematically correct, but the second version is more common in exams.

Did you know? This relationship between roots and coefficients is part of Vieta's Formulas. It works for higher-degree equations too (like cubics), but for this section of your International GCSE, we only focus on quadratics!

Key Takeaways

  • To form an equation, you only need the Sum and Product of the roots.
  • The template is always \(x^2 - (\text{Sum})x + \text{Product} = 0\).
  • If given a related equation, find \(\alpha + \beta = -b/a\) and \(\alpha\beta = c/a\) first.
  • Always double-check the signs! The "minus" in the formula often reacts with a "negative" sum to create a "plus" in the final equation.