(b) Hence, or otherwise, solve 6x² – 5x – 4 < 0
Quadratic expressions and inequalities are a core part of the GCSE Maths curriculum. In this post, we'll walk through a typical two-part exam question, breaking down each step clearly. Mastering these skills is essential for your exams with AQA, Edexcel, or OCR.
The Exam Question
- We are going to solve the following:
> (a) Factorise (6x² - 5x - - 4)
> (b) Hence, or otherwise, solve (6x² - 5x - 4 < 0)
The word "hence" is a big clue! It tells us that our answer from part (a) will make solving part (b) much quicker and easier.
Part (a): Factorising (6x² - 5x - 4)
- Factorising a quadratic in the form (ax² + bx + c) where (a ≠
- 1) (here, (a = 6)) requires a methodical approach. We are looking for two binomials: ((px + q)(rx + s)).
Step 1: Multiply (a) and (c).
Here, (a = 6) and (c = -4).
(a × c = 6 × (-4) = -24).
Step 2: Find two numbers that multiply to -24 and add to (b) (-5).
We need (m) and (n) such that (m × n = -24) and (m + n = -5).
Let's list factor pairs of 24: (1,24), (2,12), (3,8), (4,6).
Since the product is negative (-24), one number must be positive and one negative. To get a sum of -5, we need the larger number (in absolute value) to be negative.
The pair (3) and (-8) works: (3 × (-8) = -24) and (3 + (-8) = -5). Perfect!
Step 3: Rewrite the middle term using these numbers.
We split the (-5x) into (3x) and (-8x).
6x² - 5x - 4 = 6x² + 3x - 8x - 4
- Step 4: Factorise in pairs.
Group the first two terms and the last two terms, and factor out the common factor from each.
(6x² + 3x) + (-8x - - 4)
From the first group: (3x(2x + 1)).
From the second group: (-4(2x + 1)).
Notice ((2x + 1)) is a common factor! - Step 5: Factor out the common binomial.
3x(2x + - 1) - 4(2x +
- 1) = (2x + 1)(3x - 4)
✅ Answer for (a): ((2x + 1)(3x - 4))
Always check by expanding: ((2x+1)(3x-4) = 6x² - 8x + 3x - 4 = 6x² - 5x - 4). Correct!
Part (b): Solving (6x² - 5x - 4 < 0)
Now we use our factorisation. The inequality becomes:
(2x + 1)(3x - 4) < 0
This product is less than zero, meaning it is negative. For the product of two things to be negative, one must be positive and the other must be negative.
Step 1: Find the critical values.
These are the values of (x) where the expression equals zero. Set each factor to zero:
2x + 1 = 0 \quad \Rightarrow \quad x = -12
3x - 4 = 0 \quad \Rightarrow \quad x = 43
These two values split the number line into three intervals.
Step 2: Test the sign of the product in each interval.
We can use a quick sign chart or sketch. The critical values are (x = -0.5) and (x ≈ 1.333).
The intervals are:
(x < -12)
(-12 < x < 43)
- (x > 43)
Pick a test value from each interval and substitute it into the factorised form ((2x+1)(3x-4)). We only care if the result is positive or negative.
- Interval 1: Let (x = -1).
(2(-1)+1 = -1) (negative)
(3(-1)-4 = -7) (negative)
Negative × Negative = Positive. ❌ Not less than 0.
- Interval 2: Let (x = 0).
- (2(0)+1 =
- 1) (positive)
(3(0)-4 = -4) (negative)
Positive × Negative = Negative. ✅ This satisfies (< 0).
- Interval 3: Let (x = 2).
- (2(2)+1 =
- 5) (positive)
(3(2)-4 = - 2) (positive)
Positive × Positive = Positive. ❌ Not less than 0.
Step 3: Consider the endpoints.
The inequality is strictly less than ((<)), not (≤). This means we do not include the points where the expression equals zero. So, (x = -12) and (x = 43) are not included in the solution.
✅ Answer for (b): (-12 < x < 43)
This means any (x) between (-0.5) and (1.333...) (but not including those numbers) will make the original quadratic expression negative.
flowchart TD
A["Start: Solve 6x² - 5x - 4 < 0"] --> B["Factorise: (2x+1)(3x-4)"];
B --> C["Find Critical Values: x = -1/2, x = 4/3"];
C --> D["Plot on Number Line"];
D --> E["Test Interval x < -1/2: Product +ve"];
D --> F["Test Interval -1/2 < x < 4/3: Product -ve"];
D --> G["Test Interval x > 4/3: Product +ve"];
F --> H["Solution: -1/2 < x < 4/3"];
Key Takeaways for Your GCSE Exam
Master the Factorising Method: The (ac) method (multiply (a) and (c), find factors that add to (b)) is reliable for harder quadratics.
"Hence" is a Command: It almost always means you must use your previous answer to save time and gain full marks.
Solving Quadratic Inequalities:
Factorise the expression.
Find the critical values (roots).
Use a number line or sign chart to test intervals.
- Pay close attention to whether the inequality is strict ((<, >)) or inclusive ((≤, ≥)) when writing your final answer.
- Practise this process with different quadratics to build your confidence. Good luck with your revision!