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Study Guide: UK K12 GCSE/A-Level: Year 9 KS3/Pre-GCSE Mathematics - Quadratic Equations, Factorising and Formula
Source: https://www.fatskills.com/key-stage-3-ks3/chapter/uk-k12-gcse-a-level-year-9-ks3pre-gcse-mathematics-quadratic-equations-factorising-and-formula

UK K12 GCSE/A-Level: Year 9 KS3/Pre-GCSE Mathematics - Quadratic Equations, Factorising and Formula

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~8 min read

Learning Objectives

By the end of this topic, students will be able to:

  • Factorise quadratic expressions in the form of (ax^2 + bx + c) into the product of two binomials;
  • Use the factorisation method to solve quadratic equations;
  • Apply the quadratic formula to solve quadratic equations;
  • Identify and correct common misconceptions in solving quadratic equations;
  • Apply mathematical techniques to solve problems involving quadratic equations.

Core Concepts

Quadratic equations are a fundamental concept in algebra, and they can be represented in the form of (ax^2 + bx + c = 0), where (a), (b), and (c) are constants. The quadratic formula is a powerful tool for solving quadratic equations, and it is given by:

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

However, not all quadratic equations can be solved using the quadratic formula. In such cases, we can use factorisation to solve the equation. Factorisation involves expressing a quadratic expression as the product of two binomials.

Factorisation

Factorisation is a technique used to express a quadratic expression in the form of ((x + p)(x + q)), where (p) and (q) are constants. To factorise a quadratic expression, we need to find two numbers whose product is equal to the constant term and whose sum is equal to the coefficient of the linear term.

For example, consider the quadratic expression (x^2 + 5x + 6). To factorise this expression, we need to find two numbers whose product is equal to 6 and whose sum is equal to 5. These numbers are 2 and 3, so we can write the expression as:

[x^2 + 5x + 6 = (x + 2)(x + 3)]

Solving Quadratic Equations using Factorisation

To solve a quadratic equation using factorisation, we need to set the expression equal to zero and then factorise the expression. For example, consider the quadratic equation (x^2 + 5x + 6 = 0). We can solve this equation by factorising the expression as follows:

[x^2 + 5x + 6 = (x + 2)(x + 3) = 0]

This tells us that either ((x + 2) = 0) or ((x + 3) = 0). Solving for (x), we get:

[x + 2 = 0 \Rightarrow x = -2]

or

[x + 3 = 0 \Rightarrow x = -3]

Therefore, the solutions to the quadratic equation are (x = -2) and (x = -3).

Worked Examples

Example 1: Factorising a Quadratic Expression

Factorise the quadratic expression (x^2 + 7x + 12).

To factorise this expression, we need to find two numbers whose product is equal to 12 and whose sum is equal to 7. These numbers are 3 and 4, so we can write the expression as:

[x^2 + 7x + 12 = (x + 3)(x + 4)]

Example 2: Solving a Quadratic Equation using Factorisation

Solve the quadratic equation (x^2 + 7x + 12 = 0) using factorisation.

We can solve this equation by factorising the expression as follows:

[x^2 + 7x + 12 = (x + 3)(x + 4) = 0]

This tells us that either ((x + 3) = 0) or ((x + 4) = 0). Solving for (x), we get:

[x + 3 = 0 \Rightarrow x = -3]

or

[x + 4 = 0 \Rightarrow x = -4]

Therefore, the solutions to the quadratic equation are (x = -3) and (x = -4).

Example 3: Applying the Quadratic Formula

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

The quadratic formula is given by:

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

In this case, (a = 1), (b = 5), and (c = 6). Plugging these values into the formula, we get:

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

Simplifying, we get:

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

[x = \frac{-5 \pm \sqrt{1}}{2}]

[x = \frac{-5 \pm 1}{2}]

Therefore, the solutions to the quadratic equation are (x = -3) and (x = -2).

Common Misconceptions

One common misconception when solving quadratic equations is to assume that the solutions are always real numbers. However, in some cases, the solutions may be complex numbers.

For example, consider the quadratic equation (x^2 + 1 = 0). Using the quadratic formula, we get:

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

Simplifying, we get:

[x = \frac{-1 \pm \sqrt{-3}}{2}]

This tells us that the solutions are complex numbers, specifically (x = \frac{-1 + i\sqrt{3}}{2}) and (x = \frac{-1 - i\sqrt{3}}{2}).

Exam Tips

When solving quadratic equations, make sure to check your solutions by plugging them back into the original equation. This will help you avoid common mistakes and ensure that your solutions are correct.

Also, be careful when using the quadratic formula. Make sure to simplify the expression under the square root sign and then plug in the values of (a), (b), and (c).

MCQs with Explanations

MCQ 1: [F]

What is the first step in solving a quadratic equation using factorisation?

A) Set the expression equal to zero B) Factorise the expression C) Use the quadratic formula D) Solve for (x)

Correct answer: B) Factorise the expression

Why the distractors fail: Option A is incorrect because setting the expression equal to zero is the first step in solving a quadratic equation, but it is not the first step in solving it using factorisation. Option C is incorrect because the quadratic formula is used to solve quadratic equations, but it is not the first step in solving them using factorisation. Option D is incorrect because solving for (x) is the final step in solving a quadratic equation, not the first step.

MCQ 2: [H]

What is the quadratic formula?

A) (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) B) (x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}) C) (x = \frac{-b \mp \sqrt{b^2 - 4ac}}{2a}) D) (x = \frac{b \mp \sqrt{b^2 - 4ac}}{2a})

Correct answer: A) (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})

Why the distractors fail: Option B is incorrect because the correct formula has a minus sign in the numerator, not a plus sign. Option C is incorrect because the correct formula has a plus sign in the numerator, not a minus sign. Option D is incorrect because the correct formula has a plus sign in the numerator, not a minus sign.

MCQ 3: [F]

What is the difference between the quadratic formula and factorisation?

A) The quadratic formula is used to solve quadratic equations, while factorisation is used to solve linear equations. B) The quadratic formula is used to solve linear equations, while factorisation is used to solve quadratic equations. C) The quadratic formula is used to solve quadratic equations, while factorisation is used to solve quadratic equations. D) The quadratic formula is used to solve quadratic equations, while factorisation is used to solve linear equations.

Correct answer: C) The quadratic formula is used to solve quadratic equations, while factorisation is used to solve quadratic equations.

Why the distractors fail: Option A is incorrect because the quadratic formula is used to solve quadratic equations, not linear equations. Option B is incorrect because factorisation is used to solve quadratic equations, not linear equations. Option D is incorrect because the quadratic formula is used to solve quadratic equations, not linear equations.

MCQ 4: [H]

What is the value of (x) in the quadratic equation (x^2 + 5x + 6 = 0)?

A) (x = -3) B) (x = -4) C) (x = -2) D) (x = 3)

Correct answer: A) (x = -3)

Why the distractors fail: Option B is incorrect because the correct solution is (x = -3), not (x = -4). Option C is incorrect because the correct solution is (x = -3), not (x = -2). Option D is incorrect because the correct solution is (x = -3), not (x = 3).

MCQ 5: [F]

What is the first step in solving a quadratic equation using the quadratic formula?

A) Plug in the values of (a), (b), and (c) B) Simplify the expression under the square root sign C) Set the expression equal to zero D) Solve for (x)

Correct answer: B) Simplify the expression under the square root sign

Why the distractors fail: Option A is incorrect because plugging in the values of (a), (b), and (c) is the second step in solving a quadratic equation using the quadratic formula. Option C is incorrect because setting the expression equal to zero is the first step in solving a quadratic equation, but it is not the first step in solving it using the quadratic formula. Option D is incorrect because solving for (x) is the final step in solving a quadratic equation, not the first step.

Short-answer Questions

Question 1

Solve the quadratic equation (x^2 + 5x + 6 = 0) using factorisation.

Answer

To solve this equation, we need to factorise the expression. We can do this by finding two numbers whose product is equal to 6 and whose sum is equal to 5. These numbers are 2 and 3, so we can write the expression as:

[x^2 + 5x + 6 = (x + 2)(x + 3) = 0]

This tells us that either ((x + 2) = 0) or ((x + 3) = 0). Solving for (x), we get:

[x + 2 = 0 \Rightarrow x = -2]

or

[x + 3 = 0 \Rightarrow x = -3]

Therefore, the solutions to the quadratic equation are (x = -2) and (x = -3).

Question 2

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

Answer

To solve this equation, we need to plug in the values of (a), (b), and (c) into the quadratic formula. In this case, (a = 1), (b = 5), and (c = 6). Plugging these values into the formula, we get:

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

Simplifying, we get:

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

[x = \frac{-5 \pm \sqrt{1}}{2}]

[x = \frac{-5 \pm 1}{2}]

Therefore, the solutions to the quadratic equation are (x = -3) and (x = -2).

Question 3

What is the difference between the quadratic formula and factorisation?

Answer

The quadratic formula is a method for solving quadratic equations that involves plugging in the values of (a), (b), and (c) into a formula. Factorisation, on the other hand, is a method for solving quadratic equations that involves expressing the expression as the product of two binomials. While both methods can be used to solve quadratic equations, the quadratic formula is generally more efficient and easier to use than factorisation.