Quadratic Equations · Lesson 2 of 4
Solving Quadratic Equations by Factorization
“Splitting the middle term feels like performing open-heart surgery on an $x^2$ polynomial, except if you mess up, the only casualty is your sanity.”
• Understand why factorisation can solve a quadratic equation. • Factorise quadratic expressions by splitting the middle term. • Use the zero-product property to obtain roots. • Recognise distinct, equal and irrational roots from factorised forms. • Recognise when simple factorisation is not convenient.
Solution of a Quadratic Equation by Factorisation
A quadratic equation may look difficult at first, but in many cases, it can be solved by breaking it into simpler parts. Suppose a quadratic expression can be written as the product of two linear factors.
Instead of solving the whole equation at once, we set each factor equal to zero and solve them separately.
This works because of the zero-product property: If the product of two numbers is zero, then at least one of those numbers must be zero. For example, if (x − 2)(x + 3) = 0 then either: x − 2 = 0 or x + 3 = 0 So the solutions are x = 2 and x = −3.
This method is called factorisation. It turns a quadratic equation into two simpler linear equations, making the solutions much easier to find.
Problem
Solve (x − 4)(x + 3) = 0.
- 1.Either x − 4 = 0 or x + 3 = 0.
- 2.So x = 4 or x = −3.
How to Factorise ax² + bx + c
The standard method is called splitting the middle term. For ax² + bx + c, search for two numbers whose product is ac and whose sum is b. These numbers replace the middle coefficient b.
| Step | Action |
|---|---|
| 1 | Write the equation in standard form ax² + bx + c = 0. |
| 2 | Calculate ac. |
| 3 | Find numbers p and q with pq = ac and p + q = b. |
| 4 | Split bx as px + qx. |
| 5 | Group the four terms into two pairs. |
| 6 | Take common factors from each pair. |
| 7 | Factor the common binomial. |
| 8 | Set each linear factor equal to zero. |
You are not guessing randomly. You are searching for two numbers whose product is ac and whose sum is b.
Example 1: Two Distinct Positive Roots
Problem
Find the roots by factorisation.
- 1.a = 2, b = −5, c = 3, so ac = 6.
- 2.We need two numbers with product 6 and sum −5: −2 and −3.
- 3.Split the middle term: 2x² − 2x − 3x + 3 = 0.
- 4.Group: 2x(x − 1) − 3(x − 1) = 0.
- 5.Factor: (2x − 3)(x − 1) = 0.
- 6.So 2x − 3 = 0 or x − 1 = 0.
- 7.Therefore x = 3/2 or x = 1.
| Check | Value |
|---|---|
| ac | 6 |
| Required sum | −5 |
| Chosen numbers | −2 and −3 |
| Factors | (2x − 3)(x − 1) |
| Roots | 3/2 and 1 |
Example 2: One Positive and One Negative Root
Problem
Find the roots by factorisation.
- 1.a = 6, b = −1, c = −2, so ac = −12.
- 2.We need two numbers with product −12 and sum −1: 3 and −4.
- 3.6x² + 3x − 4x − 2 = 0.
- 4.3x(2x + 1) − 2(2x + 1) = 0.
- 5.(3x − 2)(2x + 1) = 0.
- 6.So x = 2/3 or x = −1/2.
| Check | Value |
|---|---|
| ac | −12 |
| Required sum | −1 |
| Chosen numbers | 3 and −4 |
| Factors | (3x − 2)(2x + 1) |
| Roots | 2/3 and −1/2 |
Example 3: Equal Roots
Problem
Find the roots by factorisation.
- 1.Split −2√6x as −√6x − √6x.
- 2.3x² − √6x − √6x + 2 = 0.
- 3.√3x(√3x − √2) − √2(√3x − √2) = 0.
- 4.(√3x − √2)² = 0.
- 5.So √3x − √2 = 0.
- 6.x = √2/√3 = √(2/3).
- 7.The factor repeats, so both roots are equal.
If the factorisation becomes (px + q)² = 0, the same root occurs twice.
A Quadratic That Looks Factorisable—but Simple Splitting Fails
Problem
Try to split the middle term using integers.
- 1.Here ac = 5.
- 2.Integer factor pairs are 1 and 5, or −1 and −5.
- 3.Their sums are 6 and −6, not −5.
- 4.So no integer pair has product 5 and sum −5.
- 5.Therefore the usual integer splitting method does not produce simple factors.
- 6.This does not mean the equation has no roots. Its roots are real but irrational, and the quadratic formula is a better method.
If no suitable pair exists, stop. Some quadratics factorise neatly; others are better handled by the quadratic formula.
Applying Roots to a Real Situation
Problem
The breadth x of a hall satisfies 2x² + x − 300 = 0. Find the dimensions if length = 2x + 1.
- 1.2x² − 24x + 25x − 300 = 0.
- 2.2x(x − 12) + 25(x − 12) = 0.
- 3.(x − 12)(2x + 25) = 0.
- 4.So x = 12 or x = −25/2.
- 5.Breadth cannot be negative, so x = 12 m.
- 6.Length = 2(12) + 1 = 25 m.
Quiz
Which principle allows us to find the roots after writing a quadratic expression as a product of two factors?
What are the roots of x² − 5x + 6 = 0?
Which is the correct factorization of 2x² + 7x + 3?
What is the solution of x² + 2x + 1 = 0?
How should the middle term of 6x² + x − 2 be split to factorize the expression?
Practice Problems
- Solve x² − 3x − 10 = 0 by factorisation.
- Solve 2x² + x − 6 = 0 by factorisation.
- Solve 100x² − 20x + 1 = 0 by factorisation.
- Find two numbers whose sum is 27 and product is 182.
- Find two consecutive positive integers whose squares add to 365.
- The altitude of a right triangle is 7 cm less than its base and the hypotenuse is 13 cm. Form and solve the quadratic equation.
- A workshop makes x articles, each costing ₹(2x + 3), and total cost is ₹90. Form and solve the quadratic equation.
Key Takeaways
• Factorisation turns one quadratic equation into two linear equations. • Look for two numbers with product ac and sum b. • Quadratics can have two distinct roots or a repeated root. • Some equations do not factorise neatly over rational numbers. • In real situations, reject roots that are physically impossible.