Quadratic Functions#
Linear Equations Revision#
Theory#
Before diving into quadratic equations, it’s worth taking a moment to review linear equations.
Linear equations are a class of polynomial equations that can be rearranged into the standard form:
where \(a\) and \(b\) are any constant numbers (with \(a \neq 0\)), and \(x\) is the variable we’re solving for. Once in this form, solving for \(x\) requires just a couple of steps:
We first isolate the \(x\) term by subtracting \(b\) from both sides, then divide by \(a\) to get \(x\) on its own.
Linear equations get their name because they create straight lines when graphed. They’re also called “first-order equations” or “equations of degree 1” because the highest power of \(x\) is 1 (since \(x\) could also be written as \(x^1\)).
Tip
When solving linear equations, always keep track of your operations and apply them to both sides of the equation consistently. This habit will serve you well for more complex equations later.
Interactive Visualization: Linear Equations#
Application#
Examples#
Example 1#
Let’s solve: \(3x + 9 = 3\)
Method 1: Converting to standard form first
Method 2: Direct approach
Note
You might notice we took the long way in Method 1 - we didn’t really have to bring it to standard form first - and you’d be right! For a simple linear equation, we could just directly solve as in Method 2 if all we care about is finding the value of \(x\).
However, for more advanced problems like quadratic equations, converting to standard form is almost always the first step to solving them. This habit will come in handy later.
Multiple Choice Questions#
Sector Specific Questions: Linear Equations Applications#
Key Takeaways#
Important
A linear equation can be written in the standard form \(ax + b = 0\) where \(a \neq 0\)
To solve, isolate the variable: \(x = -\frac{b}{a}\)
When graphed, a linear equation forms a straight line
The slope of the line equals \(-\frac{a}{b}\) in standard form, or \(m\) in slope-intercept form \(y = mx + c\)
Linear equations model situations with constant rates of change
See also
For more advanced linear equation concepts, look into:
Systems of linear equations
Linear inequalities
Linear functions and their transformations
Applications in linear programming
Quadratic Equations and their Form#
Theory#
Quadratic equations are the next step up from linear equations in the polynomial family. They have the following standard form:
where \(a\), \(b\), and \(c\) are constants (with \(a \neq 0\)), and \(x\) is our variable. What makes an equation quadratic is the presence of that \(x^2\) term — the highest power of \(x\) must be 2.
Getting quadratic equations into this standard form serves two main purposes:
It’s the starting point for solving them using standard techniques
The coefficients \(a\), \(b\), and \(c\) reveal important information about the graph of the equation
If you can transform another type of equation into quadratic form without violating any mathematical rules, you can use quadratic techniques to solve problems that might otherwise seem challenging.
Warning
A common mistake is to confuse equations that contain \(x^2\) with quadratic equations. Remember, a true quadratic equation must be able to be written in the form \(ax^2 + bx + c = 0\). Equations like \(x^2 + \frac{1}{x} = 3\) are not quadratic despite having an \(x^2\) term!
Interactive Visualization: Quadratic Function Explorer#
Application#
Examples: Identifying and Working with Quadratic Equations#
Example 1: Expanding brackets#
Is this a quadratic equation?
To find out, we’ll expand the brackets and bring everything to one side:
Yes! This is a quadratic equation where \(a=2\), \(b=7\), and \(c=0\).
Example 2: Substitution method#
Is this a quadratic equation?
At first glance, no—the highest power is 4, not 2. However, we can use a clever substitution. Let’s define \(t = x^2\) and see what happens:
We’ve transformed it into a quadratic equation in terms of \(t\), where \(a=1\), \(b=1\), and \(c=1\). After solving for \(t\), we can find \(x\) using \(t = x^2\). This substitution trick appears frequently in exams and is worth mastering.
Tip
When faced with a polynomial equation where the powers of the variable increase by multiples of 2 (like \(x^4\), \(x^2\), \(x^0\)), try substituting \(t = x^2\) to convert it to a quadratic equation.
Example 3: Trigonometric substitution#
This follows the same logic as Example 2. We can let \(t = \tan(x)\) and substitute:
Now we have a quadratic in \(t\) with \(a=1\), \(b=1\), and \(c=2\).
Example 4: Not all equations with \(x^2\) are quadratic#
Despite having an \(x^2\) term, this is not a quadratic equation. If we try to substitute \(t = \tan(x)\):
We end up with two different variables (\(t\) and \(x\)), and no way to convert between them. The presence of the trigonometric function \(\tan(x)\) alongside the polynomial term \(x^2\) prevents this from being converted to quadratic form.
Example 5: Functions with constants#
Although it involves a trigonometric term, this is a quadratic equation because \(\sin\left(\frac{\pi}{3}\right)\) evaluates to a constant (approximately 0.866). The equation has the form:
where \(a = \sin\left(\frac{\pi}{3}\right)\), \(b = 1\), and \(c = 0\).
Multiple Choice Questions#
Sector Specific Questions: Quadratic Equations Applications#
Key Takeaways#
Important
A quadratic equation has the form \(ax^2 + bx + c = 0\) where \(a \neq 0\)
Equations can often be rewritten or transformed into quadratic form
Using substitution, we can convert some higher-degree equations into quadratics
Not all equations containing \(x^2\) are quadratics, especially if they contain other non-polynomial functions of \(x\)
A trigonometric expression that doesn’t contain \(x\) as a variable acts as a constant in the equation
Factorizing Quadratic Equations by Inspection#
Theory#
Factorizing by inspection is the most intuitive approach to solving quadratic equations. At its core, factorization means breaking down an expression into simpler parts that multiply together to give the original expression.
When we have a quadratic equation in standard form:
\(ax^2 + bx + c = 0\)
Factorizing means rewriting it as:
\((px + q)(rx + s) = 0\)
Where \(p\), \(q\), \(r\), and \(s\) are constants we need to find. The beauty of this approach is that once factorized, the solutions become immediately apparent thanks to the zero product property: if a product equals zero, at least one of the factors must be zero.
Tip
Factorizing by inspection works best when:
The coefficients are “nice” integers (no messy fractions or surds)
You have strong pattern recognition skills
You need a quick solution without using formulas
If you see fractions, decimals, or square roots in your coefficients, you’re probably better off using the quadratic formula directly.
The Factorizing Process#
Let’s break down the approach:
Ensure your equation is in standard form with zero on the right side
Look for two numbers that:
Multiply to give \(ac\) (the product of the first and last coefficients)
Add up to give \(b\) (the middle coefficient)
Use these numbers to split the middle term and create your factors
Interactive Visualization: Factorization Explorer#
Application#
Example: Step by Step#
Let’s work through:
\(x^2 + 2x - 8 = 0\)
Step 1: We need to find two numbers that multiply to give \(-8\) (which is \(a \times c = 1 \times -8\)) and add up to give \(+2\) (our \(b\) value).
Step 2: Let’s list possible number pairs that multiply to give \(-8\):
\(1 \times -8 = -8\) (sum: \(1 + (-8) = -7\))
\(2 \times -4 = -8\) (sum: \(2 + (-4) = -2\))
\(-1 \times 8 = -8\) (sum: \(-1 + 8 = 7\))
\(-2 \times 4 = -8\) (sum: \(-2 + 4 = 2\)) ← This works!
Step 3: Now we’ve found our pair: \(-2\) and \(4\). Let’s use them to split the middle term: \(x^2 + 2x - 8 = x^2 - 2x + 4x - 8\)
Step 4: Group the terms in pairs and factor each pair: \(x^2 - 2x + 4x - 8 = x(x - 2) + 4(x - 2)\)
Step 5: Factor out the common term \((x - 2)\): \(x(x - 2) + 4(x - 2) = (x + 4)(x - 2)\)
So our factorized equation is: \((x + 4)(x - 2) = 0\)
Finding the Solutions#
Now that we have factorized the equation, the solutions are straightforward:
If \((x + 4)(x - 2) = 0\), then either:
\(x + 4 = 0\), which gives \(x = -4\)
\(x - 2 = 0\), which gives \(x = 2\)
We can verify these solutions:
For \(x = -4\): \((-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0\) ✓
For \(x = 2\): \((2)^2 + 2(2) - 8 = 4 + 4 - 8 = 0\) ✓
\(x = 2\) and \(x = -4\) are the roots of the quadratic equation.
Warning
When the leading coefficient isn’t 1 (i.e., \(a \neq 1\)), factorizing by inspection becomes more challenging because we have more combinations to consider.
Two approaches for these cases:
First multiply everything by \(a\) to make the coefficient of \(x^2\) equal to \(a^2\), then find factors where the outer terms multiply to give \(a^2 \times c\)
Use the quadratic formula directly (which we’ll cover in a later section)
Multiple Choice Questions#
Sector Specific Questions: Factorization Applications#
Key Takeaways#
Important
Factorizing a quadratic means writing it in the form \((px + q)(rx + s) = 0\)
The zero product property lets us set each factor equal to zero to find solutions
To factorize \(ax^2 + bx + c\), find two numbers that multiply to give \(ac\) and add to give \(b\)
Factorization works best with “nice” integer coefficients
The solutions of a quadratic equation are also called the “roots” or “zeros” of the corresponding function
Graphical Solutions to Quadratic Equations#
Theory#
Understanding the Parabola#
When we solve quadratic equations in the form \(ax^2 + bx + c = 0\), we’re searching for values of \(x\) that make the expression equal to zero. But what happens if we evaluate this expression for many different values of \(x\)?
If we plot these inputs and outputs on a coordinate plane, with \(x\) on the horizontal axis and the result on the vertical axis, we get a distinctive U-shaped curve called a parabola. This graphical representation lets us visualize quadratic equations in a powerful way:
\(y = ax^2 + bx + c\)
Where \(y\) represents the output value for any given input \(x\).
Tip
You can think of the solutions to \(ax^2 + bx + c = 0\) as the points where the parabola \(y = ax^2 + bx + c\) crosses the x-axis (where \(y = 0\)). This visual interpretation makes many quadratic equation properties immediately apparent!
Key Insights from Graphing Quadratics#
The graph of a quadratic equation reveals several important features:
X-intercepts: The points where the parabola crosses the x-axis are precisely the solutions to the quadratic equation. At these points, \(y = 0\), which means they satisfy our original equation \(ax^2 + bx + c = 0\).
Vertex (turning point): The lowest point of the parabola (if \(a > 0\)) or highest point (if \(a < 0\)) indicates where the function changes direction. If this point lies exactly on the x-axis, it means the equation has a repeated root.
Number of solutions: A parabola can:
Cross the x-axis twice (two distinct real solutions)
Touch the x-axis once (one repeated real solution)
Never cross the x-axis (two complex solutions)
Completing the Square#
Another useful way to express quadratic expressions is in “completed square form”:
\(a(x + p)^2 + q\)
Where \(a\), \(p\), and \(q\) are constants. This form provides immediate insight into the graph’s shape and position:
\(a\) determines whether the parabola opens upward (\(a > 0\)) or downward (\(a < 0\))
\(p\) tells us the horizontal shift of the vertex (it’s at \(x = -p\))
\(q\) gives us the y-coordinate of the vertex
Converting from standard form to completed square form is a technique called “completing the square,” which is also the foundation for deriving the quadratic formula.
Warning
When completing the square, always remember to:
Factor out the leading coefficient \(a\) first (if it’s not 1)
Only complete the square for the terms inside the parentheses
Balance any operations by applying them to both sides of the equation
The Process of Completing the Square#
To convert \(ax^2 + bx + c\) to \(a(x + p)^2 + q\):
Factor out the coefficient \(a\) from the first two terms
Complete the square for the expression inside the parentheses
Add and subtract the necessary constant to maintain equality
Interactive Visualization: Completing the Square#
Application#
Examples: Graphical Solutions and Completing the Square#
Example 1: When \(a = 1\)#
Let’s convert \(x^2 + 3x - 4\) to completed square form:
Step 1: Since \(a = 1\), we can proceed directly to completing the square.
Step 2: To complete the square for \(x^2 + 3x\), we:
Take half the coefficient of \(x\): \(\frac{3}{2}\)
Square it: \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\)
Add and subtract this value to maintain equality:
\(x^2 + 3x - 4 = x^2 + 3x + \frac{9}{4} - \frac{9}{4} - 4\)
Step 3: Recognize the perfect square and simplify:
\(x^2 + 3x - 4 = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} - 4 = \left(x + \frac{3}{2}\right)^2 - \frac{25}{4}\)
So the completed square form is: \(\left(x + \frac{3}{2}\right)^2 - \frac{25}{4}\)
From this form, we can immediately identify:
The vertex is at \(\left(-\frac{3}{2}, -\frac{25}{4}\right)\) or \((-1.5, -6.25)\)
Since the vertex’s y-coordinate is negative and \(a > 0\), the parabola must cross the x-axis at two points
Example 2: When \(a \neq 1\)#
Let’s convert \(5x^2 + 2x + 10\) to completed square form:
Step 1: Factor out the coefficient of \(x^2\):
\(5x^2 + 2x + 10 = 5\left(x^2 + \frac{2}{5}x\right) + 10\)
Step 2: Complete the square inside the parentheses:
Take half the coefficient of \(x\): \(\frac{1}{5}\)
Square it: \(\left(\frac{1}{5}\right)^2 = \frac{1}{25}\)
Add and subtract the necessary value inside the parentheses:
\(5\left(x^2 + \frac{2}{5}x + \frac{1}{25} - \frac{1}{25}\right) + 10\)
Step 3: Rearrange and simplify:
\(5\left(x^2 + \frac{2}{5}x + \frac{1}{25}\right) - 5 \cdot \frac{1}{25} + 10\) \(5\left(x + \frac{1}{5}\right)^2 - \frac{5}{25} + 10\) \(5\left(x + \frac{1}{5}\right)^2 - \frac{1}{5} + 10\) \(5\left(x + \frac{1}{5}\right)^2 + \frac{49}{5}\)
So the completed square form is: \(5\left(x + \frac{1}{5}\right)^2 + \frac{49}{5}\)
From this form, we can identify:
The vertex is at \(\left(-\frac{1}{5}, \frac{49}{5}\right)\) or \((-0.2, 9.8)\)
Since the vertex’s y-coordinate is positive and \(a > 0\), the parabola never crosses the x-axis, meaning the equation \(5x^2 + 2x + 10 = 0\) has no real solutions
See also
The completed square form isn’t just useful for finding the vertex of a parabola. It also:
Makes it easier to sketch the graph of a quadratic function
Provides a pathway to the quadratic formula
Helps solve problems involving optimization
Simplifies integrating certain expressions in calculus
Multiple Choice Questions#
Sector Specific Questions: Graphical Solutions Applications#
Key Takeaways#
Important
The graph of a quadratic equation is a parabola
The zeros or roots of the quadratic equation are the x-intercepts of the parabola
The vertex of the parabola represents the minimum (if \(a > 0\)) or maximum (if \(a < 0\)) value
The completed square form \(a(x + p)^2 + q\) directly reveals the vertex at \((-p, q)\)
A parabola can have two distinct real roots, one repeated real root, or no real roots, depending on whether it crosses, touches, or misses the x-axis
Deriving the Quadratic Formula#
Theory#
The quadratic formula gives us a direct way to solve any quadratic equation. While it may seem like a formula to memorize, understanding its derivation builds deeper insight into quadratic equations.
We’re aiming to derive the formula:
Starting from the standard form:
The first step is to divide by \(a\) to neaten things up:
Next, we put it into completed square form by adding and subtracting the same term:
This gives us:
Simplifying the right side:
Taking the square root of both sides:
Solving for \(x\):
And there we have the quadratic formula!
Tip
When using the quadratic formula:
Always check that your equation is in standard form \(ax^2 + bx + c = 0\)
Identify the values of \(a\), \(b\), and \(c\) correctly
Be careful with negative signs, especially in \(b\)
Always verify your solutions by substituting back into the original equation
Interactive Visualization: Quadratic Formula#
Application#
Examples: Using the Quadratic Formula#
Example 1: A Basic Application#
Solve \(2x^2 + 5x - 3 = 0\)
We identify \(a = 2\), \(b = 5\), and \(c = -3\). Using the quadratic formula:
So the solutions are \(x = \frac{1}{2}\) and \(x = -3\).
Verification:
\(2(\frac{1}{2})^2 + 5(\frac{1}{2}) - 3 = 2(\frac{1}{4}) + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 0\) ✓
\(2(-3)^2 + 5(-3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0\) ✓
Example 2: When the Discriminant is Zero#
Solve \(4x^2 - 12x + 9 = 0\)
We identify \(a = 4\), \(b = -12\), and \(c = 9\). Using the quadratic formula:
There is only one solution: \(x = \frac{3}{2}\), which is a repeated root.
This means the parabola touches the x-axis at exactly one point, \(x = \frac{3}{2}\).
Example 3: When the Discriminant is Negative#
Solve \(x^2 + x + 1 = 0\)
We identify \(a = 1\), \(b = 1\), and \(c = 1\). Using the quadratic formula:
Since \(\sqrt{-3}\) is not a real number, this equation has no real solutions. The solutions are:
These are complex numbers, and the graph of \(y = x^2 + x + 1\) never crosses the x-axis.
Warning
When the discriminant \(b^2 - 4ac\) is negative, the quadratic equation has no real solutions. In such cases, the solutions are complex numbers and the graph of the corresponding quadratic function does not intersect the x-axis.
Multiple Choice Questions#
Sector Specific Questions: Quadratic Formula Applications#
Key Takeaways#
Important
The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) gives us a direct way to find solutions for any quadratic equation
It works for all quadratic equations, even those that cannot be easily factorized
The formula is derived by completing the square on the general form \(ax^2 + bx + c = 0\)
The expression under the square root (\(b^2 - 4ac\)) is called the discriminant and determines the nature of the solutions
When using the formula, always pay attention to signs and ensure your equation is in standard form
The Discriminant: Understanding the Nature of Quadratic Solutions#
Theory#
What is the Discriminant?#
For any quadratic equation in standard form \(ax^2 + bx + c = 0\), the discriminant (denoted by \(\Delta\)) is defined as:
This expression appears under the square root in the quadratic formula:
Tip
The discriminant acts as a “solution predictor” - you can determine the number and type of solutions by evaluating this single expression, without actually solving the equation!
The Three Scenarios#
The value of the discriminant determines both the number and type of solutions:
Case 1: Positive Discriminant (\(\Delta > 0\))#
Algebraically: The equation has two distinct real solutions
Graphically: The parabola crosses the x-axis at two different points
Example: For \(x^2 - 5x + 6 = 0\):
\(\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0\)
Solutions are \(x = 2\) and \(x = 3\) (two distinct real roots)
The parabola crosses the x-axis at \(x = 2\) and \(x = 3\)
Case 2: Zero Discriminant (\(\Delta = 0\))#
Algebraically: The equation has exactly one repeated real solution
Graphically: The parabola touches the x-axis exactly once at its vertex
Example: For \(x^2 - 6x + 9 = 0\):
\(\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0\)
The only solution is \(x = 3\) (repeated twice)
The parabola’s vertex is at \((3,0)\), just touching the x-axis
Case 3: Negative Discriminant (\(\Delta < 0\))#
Algebraically: The equation has two complex solutions (not real numbers)
Graphically: The parabola never intersects the x-axis
Example: For \(x^2 + 1 = 0\):
\(\Delta = 0^2 - 4(1)(1) = -4 < 0\)
Solutions are \(x = \pm i\) (complex numbers)
The parabola sits entirely above the x-axis with vertex at \((0,1)\)
The Discriminant and Factorization#
The discriminant also tells us about factorization possibilities:
When \(\Delta > 0\): The quadratic can be factored into two different linear factors with real coefficients
When \(\Delta = 0\): The quadratic can be factored as a perfect square: \((x - r)^2\)
When \(\Delta < 0\): The quadratic cannot be factored with real coefficients
Warning
Be careful with your signs when computing the discriminant. A common mistake is to forget the negative sign when \(a\) or \(c\) is negative.
Interactive Visualization: The Discriminant#
Application#
Examples: Using the Discriminant#
Example 1: Analyzing an Equation#
For the equation \(2x^2 - 5x + 3 = 0\), calculate the discriminant and determine the nature of its solutions.
We identify \(a = 2\), \(b = -5\), and \(c = 3\).
Since \(\Delta > 0\), the equation has two distinct real solutions. We could find these using the quadratic formula:
So \(x = \frac{6}{4} = \frac{3}{2}\) or \(x = \frac{4}{4} = 1\).
Example 2: Finding a Condition for a Single Root#
For what value of \(k\) will the equation \(x^2 + kx + 4 = 0\) have exactly one real solution?
For the equation to have exactly one real solution, its discriminant must equal zero.
So \(k = 4\) or \(k = -4\) will give the equation exactly one real solution.
Example 3: No Real Solutions#
Is it possible for the equation \(2x^2 + kx + 3 = 0\) to have no real solutions? If so, for what range of values of \(k\)?
For the equation to have no real solutions, its discriminant must be negative.
So the equation will have no real solutions when \(k\) is between approximately \(-4.9\) and \(4.9\).
Multiple Choice Questions#
Sector Specific Questions: Discriminant Applications#
Key Takeaways#
Important
The discriminant \(\Delta = b^2 - 4ac\) determines the nature of solutions to a quadratic equation
When \(\Delta > 0\), the equation has two distinct real solutions
When \(\Delta = 0\), the equation has one repeated real solution
When \(\Delta < 0\), the equation has two complex solutions but no real solutions
The discriminant can be used to analyze problems without solving the full equation
Summary#
In this module, we explored quadratic equations and their various solution methods:
Important
Standard Form: \(ax^2 + bx + c = 0\) where \(a \neq 0\)
Factorization: \((px + q)(rx + s) = 0\), best for quadratics with “nice” integer coefficients
Graphical Methods: Understanding parabolas and using the completed square form \(a(x + p)^2 + q\)
Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) works for any quadratic equation
Discriminant: \(\Delta = b^2 - 4ac\) determines the number and type of solutions
Each method has its strengths:
Factorization is quick when the roots are rational numbers
Completing the square reveals the vertex and graph properties
The quadratic formula works universally, even for complex solutions
The discriminant lets you analyze the nature of solutions without solving
Mastering these techniques gives you powerful tools to solve a wide range of mathematical and real-world problems across various disciplines.