Quadratic Inequalities Revision Notes for HSC SSCE Mathematics Advanced

Quadratic Inequalities

Introduction

In this topic, you will learn how to solve quadratic inequalities using two different approaches: algebraically and graphically. Understanding quadratic inequalities is essential for finding ranges of values that satisfy certain conditions, which has many practical applications in mathematics and real-world problem-solving.

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What You'll Learn:

You will be able to:

Find solutions by identifying boundary points and testing intervals
Solve inequalities graphically by comparing functions
Represent your solutions using interval notation and number lines
Work with inequalities involving quadratic and linear functions or two quadratic functions

What is a quadratic inequality?

A quadratic inequality is a statement that compares a quadratic function to a constant or another function using an inequality symbol. The quadratic function has the form:

$f(x) = ax^2 + bx + c$

where $a$ , $b$ , and $c$ are constants and $a \neq 0$ .

Instead of finding where the function equals a particular value (as with equations), we need to find the range of x-values where the inequality is satisfied. For example:

$f(x) > k$ (function is greater than constant $k$ )
$f(x) \leq k$ (function is less than or equal to constant $k$ )
$f(x) \geq g(x)$ (one function is greater than or equal to another)

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Think of quadratic inequalities as asking: "For which $x$ -values is my quadratic function above, below, or equal to a certain line or curve?" This is fundamentally different from solving equations, where we find specific points.

Solving quadratic inequalities with constants

When a quadratic inequality compares a quadratic function to a constant value, such as $f(x) > k$ , we can solve it using either an algebraic or graphical approach. Both methods are equally valid, and you can choose the one that works best for you or verify your answer by using both.

The algebraic method

The algebraic method involves these key steps:

Step 1: Find the boundary points

Convert the inequality to an equation by replacing the inequality symbol with an equals sign. Solve $f(x) = k$ to find the boundary points where the parabola crosses the horizontal line $y = k$ .

Step 2: Test intervals

The boundary points divide the number line into intervals. Select a test value from within each interval and substitute it into the original inequality to check whether the inequality holds in that interval.

Step 3: Write the solution

Based on your testing, identify which intervals satisfy the inequality. Express your answer using interval notation or as a compound inequality.

The graphical method

The graphical approach provides a visual way to solve the inequality:

Step 1: Sketch the functions

Plot the graph of $y = f(x)$ and the horizontal line $y = k$ on the same set of axes.

Step 2: Identify the relevant regions

Look at where the graph of the parabola is positioned relative to the line:

For $f(x) > k$ : find where the parabola is above the line
For $f(x) < k$ : find where the parabola is below the line
For $f(x) \geq k$ or $f(x) \leq k$ : include the boundary points where the graphs intersect

Step 3: Read off the solution

Determine the $x$ -values corresponding to the relevant region and write your solution.

The role of the discriminant

The discriminant of the equation $f(x) = k$ is given by:

$\Delta = b^2 - 4ac$

chatImportant

The Discriminant Determines the Number of Boundary Points:

The discriminant tells us how many boundary points exist:

$\Delta > 0$ : Two distinct boundary points (parabola crosses the line twice)
$\Delta = 0$ : One boundary point (parabola touches the line at exactly one point)
$\Delta < 0$ : No boundary points (parabola is entirely above or entirely below the line)

Understanding the discriminant helps you anticipate what type of solution to expect.

Worked example 1: Solving $x^2 - 4x - 5 \leq 0$

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Worked Example: Solving a Quadratic Inequality

Let's solve the inequality $x^2 - 4x - 5 \leq 0$ using both methods.

Part a: Algebraic solution

Step 1: Convert to an equation and find boundary points

Replace the inequality symbol with an equals sign:

$x^2 - 4x - 5 = 0$

Factorise the quadratic expression:

$(x - 5)(x + 1) = 0$

Apply the null factor law. If the product of two factors equals zero, then at least one factor must equal zero:

$x - 5 = 0 \text{ or } x + 1 = 0$

$x = 5 \text{ or } x = -1$

The boundary points are at x = -1 and x = 5.

Step 2: Test intervals

The boundary points divide the number line into three intervals:

$(-\infty, -1)$
$(-1, 5)$
$(5, \infty)$

Choose a test point from each interval. Good test points are $x = -2$ , $x = 0$ , and $x = 6$ .

For each test point, substitute into the factorised form $(x - 5)(x + 1)$ and check if the result satisfies the inequality $\leq 0$ :

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Step 3: Write the solution

The inequality $(x - 5)(x + 1) \leq 0$ is satisfied in the interval $(-1, 5)$ . Since the original inequality includes "equal to" (the $\leq$ symbol), we include the boundary points.

Solution: -1 ≤ x ≤ 5

Part b: Graphical solution

Step 1: Set up the graph

We need to plot $y = x^2 - 4x - 5$ and identify where the parabola is on or below the $x$ -axis (since we're comparing to $0$ , which is represented by $y = 0$ ).

Step 2: Sketch and analyse

The parabola crosses the $x$ -axis at $(-1, 0)$ and $(5, 0)$ . These are the boundary points we found algebraically.

Step 3: Identify the solution region

The parabola $y = x^2 - 4x - 5$ is on or below the x-axis (where $y = 0$ ) between $x = -1$ and $x = 5$ .

Solution: -1 ≤ x ≤ 5

Notice that both methods give the same answer, confirming our solution.

Solving quadratic inequalities with linear or quadratic functions

Sometimes we need to solve inequalities that compare a quadratic function to a linear function or to another quadratic function. For example:

Comparing quadratic to linear: $f(x) \geq mx + d$
Comparing two quadratics: $f(x) < h(x)$ where $h(x) = a_2x^2 + b_2x + c_2$

The solution process is similar to what we've already learned, but instead of finding where a parabola crosses a horizontal line, we find where two curves intersect.

The algebraic method

Step 1: Find intersection points

Set the two functions equal to each other and solve for $x$ . These intersection points become your boundary points.

Step 2: Test intervals

Use the boundary points to divide the number line into intervals. Choose test points and substitute them into the original inequality to determine which intervals satisfy the condition.

Step 3: State the solution

Express your answer based on which intervals make the inequality true.

The graphical method

Step 1: Plot both functions

Sketch the graphs of both functions on the same coordinate plane.

Step 2: Identify the relevant region

Determine where one function is above, below, or equal to the other function, depending on the inequality symbol.

Step 3: Read the solution

Find the $x$ -values that correspond to the relevant region.

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Choosing Between Methods:

The algebraic method is often more precise and works well when factorisation is straightforward. The graphical method provides visual insight and can be particularly helpful when dealing with more complex expressions or when you want to verify your algebraic solution.

Worked example 2: Solving $3x^2 + x \geq 2x^2 + 2$

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Worked Example: Comparing Two Quadratic Functions

Let's solve this inequality comparing two quadratic functions.

Part a: Algebraic solution

Step 1: Convert to an equation

Begin with the inequality:

$3x^2 + x \geq 2x^2 + 2$

Set the functions equal to find intersection points:

$3x^2 + x = 2x^2 + 2$

Rearrange by subtracting $2x^2 + 2$ from both sides:

$x^2 + x - 2 = 0$

Factorise:

$(x + 2)(x - 1) = 0$

Apply the null factor law:

$x = -2 \text{ or } x = 1$

The boundary points are at x = -2 and x = 1.

Step 2: Test intervals

The boundary points create three intervals:

$(-\infty, -2)$
$(-2, 1)$
$(1, \infty)$

Choose test points: $x = -3$ , $x = 0$ , and $x = 2$ .

For each test point, substitute into the factorised form $(x + 2)(x - 1)$ and check if the result satisfies $\geq 0$ :

Table

Step 3: State the solution

The inequality is satisfied in the intervals $(-\infty, -2)$ and $(1, \infty)$ . Including the boundary points (since we have $\geq$ ):

Solution: x ≤ -2 or x ≥ 1

Part b: Graphical solution

Step 1: Set up the graph

We need to plot both $y = 3x^2 + x$ and $y = 2x^2 + 2$ on the same axes.

Step 2: Sketch and analyse

The two parabolas intersect at the points $(-2, 10)$ and $(1, 4)$ .

Step 3: Identify where the first parabola is above or on the second

Looking at the graph, the parabola $y = 3x^2 + x$ is above or on the parabola $y = 2x^2 + 2$ when:

$x \leq -2$ (to the left of the first intersection point)
$x \geq 1$ (to the right of the second intersection point)

Solution: x ≤ -2 or x ≥ 1

Again, both methods confirm the same solution.

Exam tips

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Common Mistakes to Avoid:

Always check whether the inequality includes equality ( $\leq$ or $\geq$ ) or not ( $<$ or $>$ ). This determines whether you include the boundary points in your solution.
When testing intervals, choose simple values that are easy to calculate, such as $0$ , $1$ , $-1$ , etc.
If you're using the graphical method, sketch your graph carefully and clearly label all intersection points.
Double-check your factorisation by expanding to ensure you haven't made an error.
Remember that the solution to a quadratic inequality is typically a range (or ranges) of values, not just discrete points.

Remember!

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Key Points to Remember:

Quadratic inequalities compare a quadratic function to a constant or another function. Solutions are ranges of x-values, not single points.
Two solution methods are available: The algebraic method (find boundary points, test intervals) and the graphical method (sketch functions, identify regions).
Boundary points are found by solving the corresponding equation. These points divide the number line into intervals that must be tested.
The discriminant determines the number of boundary points: $\Delta > 0$ gives two points, $\Delta = 0$ gives one point, and $\Delta < 0$ means no boundary points exist.
Always check whether to include boundary points. Use $\leq$ or $\geq$ to include them; use $<$ or $>$ to exclude them.

Quadratic Inequalities (HSC SSCE Mathematics Advanced): Revision Notes