GCSE Maths

Numbers
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Integers
Prime Numbers and Composite Numbers
Positive and Negative Numbers
Order of Operations
Square Numbers and Cube Numbers
The Laws of Indices
Square Roots and Cube Roots
Triangular Numbers
The Fibonacci Sequence
Factors
Multiples
The Highest Common Factor
Fractions
Adding and Subtracting Fractions
Multiplication of Fractions
Division of Fractions
Fractional Indices
Decimal Places
Fractions and Decimals
Expressing Fractions as Decimal Numbers
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Algebra
29 Topics | 1 Quiz
Algebraic Notation
Algebraic Vocabulary
Collecting Like Terms
Substituting Values into Formulae
Rearranging Equations
Algebraic Fractions
Expanding Brackets and Simplifying
Expanding Double Brackets
Factorisation
Factorising Quadratic Expressions
Factorising Quadratics with a Coefficient of x Squared Greater than 1
The Quadratic Formula
The Difference of Two Squares
Subject of Formula
Solving Linear Equations
Quadratic Equations
Completing the Square
Sequences and Nth Term
Arithmetic and Geometric Sequences
Quadratic Sequences
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Ratio, Proportion and Rates of Change
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Ratios
Scaling Ratios
Direct Proportion
Inverse Proportion
Reverse Percentages
Units and Conversions
Scale Factors
Geometry and Measures
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2D Shapes
Types of angles
Angle Properties
Angle in Parallel Lines
Triangles
Angles in Triangles
Polygons
Lines of Symmetry
Bearings
Loci and constructions
Constructing Triangles
Transformations
Enlargement
Plans and elevations
Areas and Volumes in Map Scale Problems
Probability
6 Topics
Introduction to probability
Finding the sum of probabilities
Conditional probability
Set Notation
Venn diagrams
The Complement of a Set
Statistics
2 Topics
Population and samples
Representing data
Vectors
Mensuration and Calculation
7 Topics
Area
Volume
Congruence and similarity
Pythagoras’ theorem
Sine and cosine rule
Area of a triangle
Standard units of measure
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Completing the Square

GCSE Maths Algebra Completing the Square

Completing the square is an algebraic technique used to solve quadratic equations and rewrite them in a more convenient form. This method involves transforming the quadratic equation from the standard form: ax^2 + bx + c = 0 into the form:

(x + d)^2 + e = 0

  • It’s important to remember that d=\frac{b}{2a} and e = c - \frac{b^{2}}{4a}

In doing so, we are effectively ‘completing’ the square of the quadratic expression by adding and subtracting a specific value to make it a perfect square trinomial.

A perfect square trinomial is a quadratic expression that can be factored into the product of two identical binomials. For example:

x^2 + 6x + 9

This quadratic expression can be factored into the product of two identical binomials:

(x + 3)(x + 3)

Simplifying this expression, we get:

(x + 3)^2

So, x^2 + 6x + 9 is a perfect square trinomial that can be factored into (x + 3)^2.

Completing the square is very helpful when factoring or using the quadratic formula isn’t easily applicable. It can also be used to find the highest or lowest point of a curve represented by the quadratic equation.

Solving Quadratic Equations by Completing the Square

Before we look at a worked example, let’s look through the steps for completing the square:

Step 1: Write the quadratic equation in standard form (ax^2 + bx + c = 0).

Step 2: If the coefficient of x^2 (a) is not 1, factor it out from the quadratic terms.

Step 3: Divide the coefficient of x (b) by 2, square the result, and add and subtract it within parentheses.

Step 4: Combine the terms in the parentheses to form a binomial squared (x + d)^2 and rewrite the remaining constant outside the parentheses as e.

Now, let’s look at a worked example:

Solve the quadratic equation x^2 - 6x + 5 = 0

Our goal is to rewrite this equation in the form of (x – d)^2 + e = 0, where d and e are constants. To achieve this, we’ll complete the square. Here are the steps:

1. Identify the coefficient of the linear term, which is -6 in this case.

  • As the coefficient is negative, we are aiming to rewrite the equation in the form of (x - d)^2 + e = 0 and not (x + d)^2 + e = 0.

2. Take half of this coefficient:

\frac{-6}{2} = -3.

3. Square the result from step 2:

(-3)^2 = 9.

4. Add and subtract the result from step 3 to the left side of the equation:

x^2 - 6x + 9 - 9 + 5 = 0

5. Factor the first three terms into a perfect square:

(x - 3)^2 - 4 = 0

Now, the equation is in the form of (x - d)^2 + e = 0, where d = 3 and e = –4.

To solve for x, you can either use the square root property or the quadratic formula.

Using the square root property:

(x - 3)^2 - 4 = 0

(x - 3)^2 = 4

x - 3 = \pm \sqrt{4} x

So, the solutions are x = 3 \pm 2

x = 3 + 2 = 5

x = 3 - 2 = 1

Therefore, the solutions for the quadratic equation x^2 - 6x + 5 = 0 are x = 1 and x = 5.

The process is slightly different when the coefficient of the x^2 term is not 1. This is because you have to factor the coefficient a of the x^2 term out of the quadratic and linear terms, leaving you with a(x^2+\frac{b}{a}x)+c.

Examples

Example 1:

Solve the equation x^2 + 6x + 7 = 0

Start by grouping the quadratic and linear terms:

(x^2 + 6x) + 7 = 0

Complete the square by adding and subtracting the square of half of the linear coefficient:

(x^2 + 6x + 9) - 9 + 7 = 0

Rewrite as a binomial squared plus a constant: (x + 3)^2 - 2 = 0

Now, isolate x: (x + 3)^2 = 2

x + 3 = \pm \sqrt{2}

x = -3 \pm \sqrt{2}

Example 2:

Solve the equation 2x^2 - 8x + 5 = 0

Factor out the coefficient of x^2:

2(x^2 - 4x) + 5 = 0

Complete the square:

2(x^2 - 4x + 4) - 8 + 5 = 0

Rewrite as a binomial squared plus a constant:

2(x - 2)^2 - 3 = 0

Isolate x:

2(x - 2)^2 = 3

(x - 2)^2 = \frac{3}{2}

x - 2 = \pm \sqrt{\frac{3}{2}}

x = 2 \pm \sqrt{\frac{3}{2}}

Example 3:

Write the expression 2x^{2} + 8x + 10 in the form a(x + b)^{2} + c, where a, b and c are constants.

1. Factor the coefficient of x² (which is 2) out of the quadratic and linear terms:

2(x^2 + 4x) + 10

2. Complete the square for the expression inside the parentheses:

To complete the square, we need to find a value that can be added and subtracted inside the parentheses. We’ll take half of the coefficient of x, which is 4, and square it: (4/2)² = 2² = 4.

Now we add and subtract this value inside the parentheses:

2(x^2 + 4x + 4 - 4) + 10

3. Rewrite the quadratic expression inside the parentheses as a square of a binomial:

2((x + 2)^{2} - 4) + 10

4. Distribute the coefficient of x² (2) back in:

2(x + 2)^{2} - 8 + 10

5. Simplify the constant term:

2(x + 2)^{2} + 2

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