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
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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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The Quadratic Formula

GCSE Maths Algebra The Quadratic Formula

The quadratic formula is:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. The symbol “±” indicates that there are two possible solutions for x, one involving the addition of the square root term and the other involving its subtraction.

Deriving the Quadratic Formula

The quadratic formula is derived by completing the square, a technique that involves rewriting a quadratic equation in the form of a perfect square trinomial. Let’s start with a general quadratic equation in the form:

ax^2 + bx + c = 0

Step 1: Divide both sides of the equation by a:

x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Step 2: Complete the square by adding and subtracting the square of half the coefficient of the linear term, which is \frac{b}{a}, not b. Therefore, the correct value to use here is \frac{(\frac{b}{a})}{2}. This helps create a perfect square trinomial.

To find half the coefficient of the linear term, divide the coefficient \frac{b}{a} by 2, which is the same as multiplying it by \frac{1}{2}:

\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}

Now, we’ll add and subtract the square of this value, \left(\frac{b}{2a}\right)^2:

x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0

Step 3: Combine the terms to form a perfect square trinomial on the left side:

\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0

Step 4: Isolate the square term by adding \frac{b^2}{4a^2} - \frac{c}{a} to both sides:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}

Step 5: Combine the terms on the right side:

To combine the terms on the right side of the equation, we need a common denominator. The common denominator for the fractions is 4a^2, so we multiply \frac{c}{a}</mark> by 4a^2:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}

Now, we can combine the terms:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Step 6: Take the square root of both sides and solve for x:

To solve for x, we need to take the square root of both sides of the equation:

\sqrt{\left(x + \frac{b}{2a}\right)^2} = \sqrt{\frac{b^2 - 4ac}{4a^2}}

This simplifies to:

x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}

Now, we can isolate x by subtracting \frac{b}{2a} from both sides:

x = -\frac{b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}

We can then factor out \frac{1}{2a} to form the quadratic formula:

x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}

Solving Quadratic Equations using the Quadratic Formula

To solve a quadratic equation using the quadratic formula, follow these steps:

Step 1: Identify the coefficients a, b, and c in the equation.

Step 2: Substitute the coefficients into the quadratic formula.

Step 3: Simplify the expression under the square root (if possible).

Step 4: Calculate the two possible values of x.

Let’s look at an example:

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

1. Identify the coefficients: a = 1, b = -5, and c = 6.

2. Substitute the coefficients into the quadratic formula:

x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}

3. Simplify the expression under the square root:

x = \frac{5 \pm \sqrt{25 - 24}}{2}

4. Calculate the two possible values of x:

x = \frac{5 \pm \sqrt{1}}{2}

x_1 = \frac{5 + 1}{2} = 3

x_2 = \frac{5 - 1}{2} = 2

So, the solutions to the quadratic equation x^2 - 5x + 6 = 0 are x_1 = 3 and x_2 = 2.

Let’s look at some more examples:

Examples

Question 1:

Solve the quadratic equation: 5x^2 + 3x - 2 = 0

In this case, a = 5, b = 3, and c = -2. Plugging these values into the quadratic formula, we get:

x = \frac{-3 \pm \sqrt{(3)^2 - 4(5)(-2)}}{2(5)}

x = \frac{-3 \pm \sqrt{9 + 40}}{10}

x = \frac{-3 \pm \sqrt{49}}{10}

So the two solutions for x are:

x = \frac{-3 + 7}{10} = \frac{4}{10} = \frac{2}{5} and x = \frac{-3 - 7}{10} = -1

Question 2:

Solve the quadratic equation: 3x^2 - 2x - 1 = 0

In this case, a = 3, b = -2, and c = -1. Plugging these values into the quadratic formula, we get:

x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)}

x = \frac{2 \pm \sqrt{4 + 12}}{6}

x = \frac{2 \pm \sqrt{16}}{6}

So the two solutions for x are:

x = \frac{2 + 4}{6} = 1 and x = \frac{2 - 4}{6} = -\frac{1}{3}

Question 3:

Solve the quadratic equation: 2x^2 - 4x - 6 = 0

In this case, a = 2, b = -4, and c = -6. Plugging these values into the quadratic formula, we get:

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}

x = \frac{4 \pm \sqrt{16 + 48}}{4}

x = \frac{4 \pm \sqrt{64}}{4}

So the two solutions for x are:

x = \frac{4 + 8}{4} = 3 and x = \frac{4 - 8}{4} = -1

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