Solution of a Quadratic Equation by Factorisation: Step-by-Step Guide

The factorisation method is a fundamental technique for solving quadratic equations. It is efficient, straightforward, and widely used in algebra. This method allows us to express a quadratic equation as a product of two linear factors and find the values of the variable by applying the Zero Product Property.

What is a Quadratic Equation Root?

A root of a quadratic equation is a value of $\boldsymbol{x}$ that satisfies the equation $\boldsymbol{ax^2 + bx + c = 0}$ . In general, for $\boldsymbol{\alpha}$ to be a root of $\boldsymbol{ax^2 + bx + c = 0}$ :

$\boldsymbol{a \cdot \alpha^2 + b \cdot \alpha + c = 0, \quad a \neq 0}$

The roots of the equation are also called the zeroes of the quadratic polynomial $\boldsymbol{ax^2 + bx + c}$ .

What is Factorisation?

Factorisation involves rewriting a quadratic equation of the form:

$\boldsymbol{ax^2 + bx + c = 0}$

as a product of two linear factors:

$\boldsymbol{(px + q)(rx + s) = 0}$

Where:

$\boldsymbol{p}$ and $\boldsymbol{r}$ are coefficients of $\boldsymbol{x}$ ,
$\boldsymbol{q}$ and $\boldsymbol{s}$ are constants.

Once the equation is factorised, the Zero Product Property is used:
If $\boldsymbol{ab = 0}$ , then $\boldsymbol{a = 0}$ or $\boldsymbol{b = 0}$ .

This property enables us to find the values of $\boldsymbol{x}$ that satisfy the equation.

Steps to Solve Quadratic Equations by Factorisation

Write the Equation in Standard Form
Ensure the quadratic equation is written as $\boldsymbol{ax^2 + bx + c = 0}$ , where $\boldsymbol{a \neq 0}$ .
Find the Factors
Identify two numbers that:
- Multiply to $\boldsymbol{a \cdot c}$ (the product of the first and last coefficients), and
- Add up to $\boldsymbol{b}$ (the middle term coefficient).
Split the Middle Term
Rewrite the middle term using the two numbers found in Step 2.
Group and Factorise
Group the terms into two pairs and factor out the greatest common factor (GCF) from each pair.
Apply the Zero Product Property
Set each factor equal to zero and solve for $\boldsymbol{x}$ .

Detailed Examples

Example 1: Solve $\boldsymbol{x^2 - 7x + 12 = 0}$

Solution:

Write in standard form.
The equation is already in standard form: $\boldsymbol{x^2 - 7x + 12 = 0}$ .
Find two numbers that multiply to $\boldsymbol{12}$ and add to $\boldsymbol{-7}$ .
The numbers are $\boldsymbol{-3}$ and $\boldsymbol{-4}$ .
Split the middle term.
$\boldsymbol{x^2 - 3x - 4x + 12 = 0}$ .
Group and factorise.
Group terms: $\boldsymbol{(x^2 - 3x) - (4x - 12) = 0}$ .
Factorise: $\boldsymbol{x(x - 3) - 4(x - 3) = 0}$ .
Factorise completely.
$\boldsymbol{(x - 3)(x - 4) = 0}$ .
Solve.
Set each factor to zero:
$\boldsymbol{x - 3 = 0 \quad \Rightarrow \quad x = 3}$ ,
$\boldsymbol{x - 4 = 0 \quad \Rightarrow \quad x = 4}$ .

Answer: $\boldsymbol{x = 3, x = 4}$ .

Example 2: Solve $\boldsymbol{2x^2 + 7x + 3 = 0}$

Solution:

Write in standard form.
The equation is already in standard form: $\boldsymbol{2x^2 + 7x + 3 = 0}$ .
Find two numbers that multiply to $\boldsymbol{2 \cdot 3 = 6}$ and add to $\boldsymbol{7}$ .
The numbers are $\boldsymbol{6}$ and $\boldsymbol{1}$ .
Split the middle term.
$\boldsymbol{2x^2 + 6x + x + 3 = 0}$ .
Group and factorise.
Group terms: $\boldsymbol{(2x^2 + 6x) + (x + 3) = 0}$ .
Factorise: $\boldsymbol{2x(x + 3) + 1(x + 3) = 0}$ .
Factorise completely.
$\boldsymbol{(2x + 1)(x + 3) = 0}$ .
Solve.
Set each factor to zero:
$\displaystyle\boldsymbol{2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}}$ ,
$\boldsymbol{x + 3 = 0 \quad \Rightarrow \quad x = -3}$ .

Answer: $\displaystyle\boldsymbol{x = -\frac{1}{2}, x = -3}$ .

Example 3: Solve $\boldsymbol{3x^2 - 8x + 4 = 0}$

Solution:

Write in standard form.
The equation is already in standard form: $\boldsymbol{3x^2 - 8x + 4 = 0}$ .
Find two numbers that multiply to $\boldsymbol{3 \cdot 4 = 12}$ and add to $\boldsymbol{-8}$ .
The numbers are $\boldsymbol{-6}$ and $\boldsymbol{-2}$ .
Split the middle term.
$\boldsymbol{3x^2 - 6x - 2x + 4 = 0}$ .
Group and factorise.
Group terms: $\boldsymbol{(3x^2 - 6x) - (2x - 4) = 0}$ .
Factorise: $\boldsymbol{3x(x - 2) - 2(x - 2) = 0}$ .
Factorise completely.
$\boldsymbol{(3x - 2)(x - 2) = 0}$ .
Solve.
Set each factor to zero:
$\displaystyle\boldsymbol{3x - 2 = 0 \quad \Rightarrow \quad x = \frac{2}{3}}$ ,
$\boldsymbol{x - 2 = 0 \quad \Rightarrow \quad x = 2}$ .

Answer: $\displaystyle\boldsymbol{x = \frac{2}{3}, x = 2}$ .

Example 4: Solve $\boldsymbol{2x^2 - 5x + 3 = 0}$

Solution:

Split the middle term $\boldsymbol{-5x}$ into $\boldsymbol{-2x}$ and $\boldsymbol{-3x}$ .
$\boldsymbol{2x^2 - 5x + 3 = 2x^2 - 2x - 3x + 3}$ .
Group the terms:
$\boldsymbol{(2x^2 - 2x) - (3x - 3)}$ .
Factorise:
$\boldsymbol{2x(x - 1) - 3(x - 1) = 0}$ .
Combine terms:
$\boldsymbol{(2x - 3)(x - 1) = 0}$ .
Solve each factor:
$\displaystyle\boldsymbol{2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2}}$ ,
$\boldsymbol{x - 1 = 0 \quad \Rightarrow \quad x = 1}$ .

Answer: $\displaystyle\boldsymbol{x = \frac{3}{2}, x = 1}$ .

Example 5: Solve $\boldsymbol{6x^2 - x - 2 = 0}$

Solution:

Split the middle term $\boldsymbol{-x}$ into $\boldsymbol{3x}$ and $\boldsymbol{-4x}$ .
$\boldsymbol{6x^2 - x - 2 = 6x^2 + 3x - 4x - 2}$ .
Group the terms:
$\boldsymbol{(6x^2 + 3x) - (4x + 2)}$ .
Factorise:
$\boldsymbol{3x(2x + 1) - 2(2x + 1) = 0}$ .
Combine terms:
$\boldsymbol{(3x - 2)(2x + 1) = 0}$ .
Solve each factor:
$\displaystyle\boldsymbol{3x - 2 = 0 \quad \Rightarrow \quad x = \frac{2}{3}}$ ,
$\displaystyle\boldsymbol{2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}}$ .

Answer: $\displaystyle\boldsymbol{x = \frac{2}{3}, x = -\frac{1}{2}}$ .

Example 6: Solve $\boldsymbol{3x^2 - 2\sqrt{6}x + 2 = 0}$

Solution:

Rewrite the equation in standard form.
Factorise:
$\boldsymbol{3x^2 - 2\sqrt{6}x + 2 = (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2})}$ .
The roots of the equation are:
$\displaystyle\boldsymbol{\sqrt{3}x - \sqrt{2} = 0 \quad \Rightarrow \quad x = \frac{\sqrt{2}}{\sqrt{3}}}$ .
This root is repeated because the factor $\boldsymbol{(\sqrt{3}x - \sqrt{2})}$ appears twice.

Answer: $\displaystyle\boldsymbol{x = \frac{\sqrt{2}}{\sqrt{3}} , \text{(repeated root)}}$ .

Example 7: Dimensions of a Prayer Hall

Find the dimensions of a hall if the quadratic equation governing its breadth is:
$\boldsymbol{2x^2 - 24x + 25 = 0}$ .

Solution:

Factorise:
$\boldsymbol{2x^2 - 24x + 25 = 2(x - 12)(x - 12.5)}$ .
Solve for $\boldsymbol{x}$ :
$\boldsymbol{x - 12 = 0 \quad \Rightarrow \quad x = 12}$ .

Since the breadth cannot be negative, $\boldsymbol{x = 12}$ . The length is $\boldsymbol{2x + 1}$ :
$\boldsymbol{2(12) + 1 = 25}$ .

Answer: Breadth = $\boldsymbol{12 , \text{m}}$ , Length = $\boldsymbol{25 , \text{m}}$ .

Practice Questions

Solve $\boldsymbol{x^2 + 5x + 6 = 0}$ .
Answer: $\boldsymbol{x = -2, x = -3}$ .
Solve $\boldsymbol{4x^2 - 4x - 8 = 0}$ .
Answer: $\boldsymbol{x = 2, x = -1}$ .

Solve $\boldsymbol{5x^2 - 13x + 6 = 0}$ .
Answer: $\displaystyle\boldsymbol{x = 2, x = \frac{3}{5}}$ .

FAQs

What if the quadratic equation cannot be factorised?admin2024-12-18T11:18:24+05:30

Solution of a Quadratic Equation by Factorisation

What is a Quadratic Equation Root?

What is Factorisation?

Steps to Solve Quadratic Equations by Factorisation

Detailed Examples

Example 1: Solve

Example 2: Solve

Example 3: Solve

Example 4: Solve

Example 5: Solve

Example 6: Solve

Example 7: Dimensions of a Prayer Hall

Practice Questions

FAQs

Related Topics

Related Posts

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Example 1: Solve $\boldsymbol{x^2 - 7x + 12 = 0}$

Example 2: Solve $\boldsymbol{2x^2 + 7x + 3 = 0}$

Example 3: Solve $\boldsymbol{3x^2 - 8x + 4 = 0}$

Example 4: Solve $\boldsymbol{2x^2 - 5x + 3 = 0}$

Example 5: Solve $\boldsymbol{6x^2 - x - 2 = 0}$

Example 6: Solve $\boldsymbol{3x^2 - 2\sqrt{6}x + 2 = 0}$