GCSE Maths

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Algebra
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Algebraic Notation
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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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Ratios
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Probability
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Introduction to Probability
Relative Frequency
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Probability Trees
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Set Notation
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Population and samples
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Mensuration and Calculation
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Area
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Congruence and similarity
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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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