Solving linear and quadratic inequalities, representing solutions on a number line and using set notation.
Answer at least 3 of 3 correctly to complete this section.
At GCSE you solved equations to find exact values. But real-world problems often ask “when is this profitable?” or “what range of speeds is safe?” — that’s inequalities. Instead of a single answer, you get a range of values.
At A-Level, quadratic inequalities are new and combine your factorising and graph-sketching skills. The method is systematic: factorise, sketch, read off the answer.
Linear inequalities work just like linear equations — with one crucial difference.
You can add, subtract, multiply, or divide both sides by a positive number, and the inequality stays the same:
2x+3>7⟹2x>4⟹x>2
But if you multiply or divide by a negative number, you must flip the inequality sign:
−3x>12⟹x<−4(sign flips!)
Why does the sign flip? Think about it on the number line. If −3x>12, then −3x is to the right of 12. When you divide by −3 you reverse the direction: small negatives become large positives and vice versa. So the direction of the inequality reverses.
Solve 5−x≤2
5−x≤2 −x≤2−5 −x≤−3 x≥3(divided by −1, so flip!)
Solutions to inequalities are shown on a number line:
Choose an inequality to see the solving steps and its number line.
Example 1: Solve 3x−1<8
3x−1<8 3x<9 x<3
Answer at least 2 of 2 correctly to complete this section.
Now here is the crucial twist. The next examples involve dividing by a negative number.
Example 2: Solve 4−3x≥−5
4−3x≥−5 −3x≥−5−4 −3x≥−9 x≤3(divided by −3, flip!)
Example 3: Solve 2x+1>3
2x+1>3 x+1>6 x>5
Answer at least 3 of 4 correctly to complete this section.
Quadratic inequalities are not solved by rearranging like linear ones. The method is:
Critical insight: A U-shaped parabola (x2 positive) is below the x-axis between its roots, and above the x-axis outside its roots.
Step 1: Factorise: x2−5x+6=(x−2)(x−3)
Step 2: Roots: x=2 and x=3
Step 3: Sketch: positive x2 coefficient, so U-shape opening upwards
Step 4: We need >0 (above the x-axis). The parabola is above the axis outside the roots:
x<2orx>3
Step 1: Factorise: x2−4=(x−2)(x+2)
Step 2: Roots: x=−2 and x=2
Step 3: Sketch: U-shape
Step 4: We need <0 (below the x-axis). The parabola dips below between the roots:
−2<x<2
Common mistake: Students solve x2>4 and write only x>2. But x2>4 also holds when x<−2. Always sketch — that’s the whole point of the method.
Solve 6+x−x2>0
Rearrange with x2 positive: −(x2−x−6)>0, which is the same as x2−x−6<0.
Factorise: (x−3)(x+2)<0
Roots: x=−2 and x=3. U-shape, need below the axis:
−2<x<3
Alternatively, factorise directly as −(x−3)(x+2), sketch the n-shape (opens downward), and read off where it is above the x-axis: between the roots, giving the same answer.
The parabola y=(x−1)(x−5) is a U-shape crossing the x-axis at x=1 and x=5. Use this to solve (x−1)(x−5)>0.
Think about it: the parabola is above the x-axis (i.e. >0) when x is outside the roots. So the answer is x<1 or x>5.
This is the method in a nutshell — factorise, sketch, read off. Now try the interactive tool below.
Choose a quadratic inequality, then step through the method.
Answer at least 3 of 4 correctly to complete this section.
At A-Level, you are expected to write solutions in set notation:
{x:x>3}reads as "the set of all x such that x>3"
For quadratic inequalities with two separate regions, use the union symbol ∪:
{x:x<−2}∪{x:x>3}
Some textbooks also use interval notation:
| Inequality | Interval notation |
|---|---|
| x>3 | (3,∞) |
| x≥3 | [3,∞) |
| x<3 | (−∞,3) |
| −2<x<3 | (−2,3) |
| −2≤x≤3 | [−2,3] |
Round brackets = endpoint not included. Square brackets = endpoint included.
Tip: ∞ always gets a round bracket — you can never “reach” infinity.
This is a crucial distinction:
“And” (intersection ∩): Both conditions must hold simultaneously. The solution is where the regions overlap.
Solve x>1 and x<5
Solution: 1<x<5 — the overlap of both regions.
“Or” (union ∪): At least one condition holds. The solution is both regions combined.
From x2−5x+6>0: x<2 or x>3
Solution: {x:x<2}∪{x:x>3}
Common mistake: Students write x<2 and x>3 for two separate regions. No number is simultaneously less than 2 and greater than 3! It must be or.
Sometimes you solve two simultaneous inequalities to find the overlap:
Find the values of x satisfying both 2x+1>5 and 3x−2<13.
From 2x+1>5: x>2
From 3x−2<13: x<5
Both must hold: 2<x<5
Answer at least 3 of 4 correctly to complete this section.
Solve 23x−5≥7
3x−5≥14 3x≥19 x≥319
In set notation: {x:x≥319}
Solve 2x2+x−3≤0
Factorise: 2x2+x−3=(2x+3)(x−1)
Roots: x=−23 and x=1
Sketch: Positive x2 coefficient means U-shape. We need ≤0 (on or below the axis), which is between the roots:
−23≤x≤1
The length of a rectangle is (x+3) cm and the width is (x−1) cm. The area must be less than 21cm2. Given x>1, find the range of values of x.
Area: (x+3)(x−1)<21
x2+2x−3<21 x2+2x−24<0 (x+6)(x−4)<0
From the sketch: −6<x<4
But we also need x>1 (width must be positive). Combining: 1<x<4.
The equation x2+kx+9=0 has no real roots. Find the range of values of k.
No real roots means b2−4ac<0:
k2−4(1)(9)<0 k2−36<0 k2<36 −6<k<6
This is itself a quadratic inequality! We used the sketch method to find the range.
Solve 1<2x−3≤9
Split into two inequalities and solve each:
1<2x−3⟹4<2x⟹2<x 2x−3≤9⟹2x≤12⟹x≤6
Combine: 2<x≤6
Answer at least 4 of 5 correctly to complete this section.
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