Quadratic Functions and Complex Numbers

Anatomy of the Parabola

A parabola is the graph of a quadratic relationship. In Grade 12, you often see parabolas when studying quadratic functions such as \(y=ax^2+bx+c\). To fully analyze a parabola, you should be able to identify its vertex, axis of symmetry, focus, directrix, and intercepts.

These features help you understand the shape, position, and important points of the graph. They also help in graphing, solving optimization problems, and connecting algebra to geometry.

1. What is a parabola?

A parabola is the set of all points that are the same distance from a fixed point and a fixed line.

• The fixed point is called the focus.
• The fixed line is called the directrix.

This geometric definition creates the familiar U-shaped or upside-down U-shaped curve.

2. Common algebraic forms of a parabola

The most common function form is the standard form:

$$y=ax^2+bx+c$$

Here:

• \(a\) tells whether the parabola opens upward or downward and how wide or narrow it is.
• \(b\) and \(c\) affect the location of the graph.

Another very useful form is the vertex form:

$$y=a(x-h)^2+k$$

In this form, the vertex is immediately visible as \((h,k)\).

For finding the focus and directrix, it is especially helpful to use the conic form of a vertical parabola:

$$ (x-h)^2 = 4p(y-k) $$

or of a horizontal parabola:

$$ (y-k)^2 = 4p(x-h) $$

In these forms:

• \((h,k)\) is the vertex.
• \(p\) is the directed distance from the vertex to the focus.
• If \(p>0\), the parabola opens up or right.
• If \(p<0\), the parabola opens down or left.

3. Main parts of a parabola

Vertex

The vertex is the turning point of the parabola. It is the lowest point if the parabola opens upward, and the highest point if it opens downward.

For \(y=a(x-h)^2+k\), the vertex is:

$$ (h,k) $$

For \(y=ax^2+bx+c\), the x-coordinate of the vertex is:

$$ x=-\frac{b}{2a} $$

Then substitute this x-value into the equation to get the y-coordinate.

Axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex and divides the parabola into two mirror-image halves.

For a vertical parabola, the axis of symmetry is:

$$ x=h $$

or, if the equation is in standard form,

$$ x=-\frac{b}{2a} $$

Intercepts

Intercepts are points where the graph crosses the axes.

• The y-intercept happens when \(x=0\).
• The x-intercepts happen when \(y=0\).

To find the y-intercept, substitute \(x=0\) into the equation.

To find x-intercepts, solve:

$$ ax^2+bx+c=0 $$

This can be done by factoring, completing the square, or using the quadratic formula:

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

Focus and directrix

The focus and directrix come from the geometric definition of the parabola.

For a vertical parabola written as

$$ (x-h)^2=4p(y-k) $$

• Vertex: \((h,k)\)
• Focus: \((h,k+p)\)
• Directrix: \(y=k-p\)
• Axis of symmetry: \(x=h\)

For a horizontal parabola written as

$$ (y-k)^2=4p(x-h) $$

• Vertex: \((h,k)\)
• Focus: \((h+p,k)\)
• Directrix: \(x=h-p\)
• Axis of symmetry: \(y=k\)

4. How the coefficient \(a\) affects the parabola

• If \(a>0\), the parabola opens upward.
• If \(a<0\), the parabola opens downward.
• If \(|a|\) is larger, the parabola is narrower.
• If \(|a|\) is smaller, the parabola is wider.

This helps you predict the graph even before calculating all of its features.

5. Converting standard form to vertex form

When a quadratic is given in standard form, the easiest way to see its structure is often to rewrite it in vertex form by completing the square.

Example pattern:

$$ y=ax^2+bx+c $$

Factor out \(a\) from the \(x^2\) and \(x\) terms, then complete the square inside the parentheses.

This process helps you find the vertex and then connect the function form to the focus and directrix.

Worked Example 1: Identify basic features from vertex form

Find the vertex, axis of symmetry, and intercepts of

$$ y=(x-3)^2-4 $$

Step 1: Find the vertex

The equation is already in vertex form \(y=a(x-h)^2+k\).

So the vertex is:

$$ (h,k)=(3,-4) $$

Step 2: Find the axis of symmetry

The axis of symmetry is:

$$ x=3 $$

Step 3: Find the y-intercept

Set \(x=0\):

$$ y=(0-3)^2-4=9-4=5 $$

So the y-intercept is:

$$ (0,5) $$

Step 4: Find the x-intercepts

Set \(y=0\):

$$ 0=(x-3)^2-4 $$

$$ 4=(x-3)^2 $$

$$ x-3=\pm 2 $$

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

So the x-intercepts are:

$$ (5,0) \text{ and } (1,0) $$

Conclusion:

• Vertex: \((3,-4)\)
• Axis of symmetry: \(x=3\)
• x-intercepts: \((1,0)\) and \((5,0)\)
• y-intercept: \((0,5)\)

Worked Example 2: Find features from standard form

Analyze the parabola

$$ y=2x^2-8x+3 $$

Step 1: Find the axis of symmetry

Use:

$$ x=-\frac{b}{2a} $$

Here, \(a=2\) and \(b=-8\).

$$ x=-\frac{-8}{2(2)}=\frac{8}{4}=2 $$

So the axis of symmetry is:

$$ x=2 $$

Step 2: Find the vertex

Substitute \(x=2\) into the equation:

$$ y=2(2)^2-8(2)+3=8-16+3=-5 $$

So the vertex is:

$$ (2,-5) $$

Step 3: Determine opening

Since \(a=2>0\), the parabola opens upward.

Step 4: Find the y-intercept

Set \(x=0\):

$$ y=2(0)^2-8(0)+3=3 $$

So the y-intercept is:

$$ (0,3) $$

Step 5: Find the x-intercepts

Set \(y=0\):

$$ 2x^2-8x+3=0 $$

Use the quadratic formula:

$$ x=\frac{-(-8)\pm\sqrt{(-8)^2-4(2)(3)}}{2(2)} $$

$$ x=\frac{8\pm\sqrt{64-24}}{4} $$

$$ x=\frac{8\pm\sqrt{40}}{4} $$

$$ x=\frac{8\pm 2\sqrt{10}}{4} $$

$$ x=2\pm \frac{\sqrt{10}}{2} $$

So the x-intercepts are:

$$ \left(2+\frac{\sqrt{10}}{2},0\right) \text{ and } \left(2-\frac{\sqrt{10}}{2},0\right) $$

Conclusion:

• Vertex: \((2,-5)\)
• Axis of symmetry: \(x=2\)
• Opens upward
• y-intercept: \((0,3)\)
• x-intercepts: \(\left(2\pm\frac{\sqrt{10}}{2},0\right)\)

Worked Example 3: Find focus and directrix

Find the vertex, focus, directrix, and axis of symmetry of

$$ (x+1)^2=8(y-2) $$

Step 1: Compare with the standard conic form

For a vertical parabola,

$$ (x-h)^2=4p(y-k) $$

Given:

$$ (x+1)^2=8(y-2) $$

This means:

• \(h=-1\)
• \(k=2\)
• \(4p=8\), so \(p=2\)

Step 2: Find the vertex

$$ (-1,2) $$

Step 3: Find the focus

For a vertical parabola, the focus is \((h,k+p)\).

$$ (-1,2+2)=(-1,4) $$

Step 4: Find the directrix

The directrix is:

$$ y=k-p=2-2=0 $$

Step 5: Find the axis of symmetry

$$ x=h=-1 $$

Conclusion:

• Vertex: \((-1,2)\)
• Focus: \((-1,4)\)
• Directrix: \(y=0\)
• Axis of symmetry: \(x=-1\)

Worked Example 4: Rewrite and then analyze

Find the vertex, axis of symmetry, and focus/directrix of

$$ x^2-4x-8y+12=0 $$

Step 1: Rearrange the equation

Group the x-terms:

$$ x^2-4x=8y-12 $$

Step 2: Complete the square

Take half of \(-4\), which is \(-2\), and square it to get \(4\).

Add 4 to both sides:

$$ x^2-4x+4=8y-12+4 $$

$$ (x-2)^2=8y-8 $$

Factor the right side:

$$ (x-2)^2=8(y-1) $$

Step 3: Compare with conic form

$$ (x-h)^2=4p(y-k) $$

So:

• \(h=2\)
• \(k=1\)
• \(4p=8\), so \(p=2\)

Step 4: State the features

• Vertex: \((2,1)\)
• Axis of symmetry: \(x=2\)
• Focus: \((2,3)\)
• Directrix: \(y=-1\)

6. Quick guide: how to identify parts from different forms

If the parabola is given in vertex form \(y=a(x-h)^2+k\):

• Vertex is \((h,k)\).
• Axis of symmetry is \(x=h\).
• Find intercepts by substitution.
• To find focus/directrix, first rewrite into conic form if needed.

If the parabola is given in standard form \(y=ax^2+bx+c\):

• Find axis using \(x=-\frac{b}{2a}\).
• Find vertex by substitution.
• Find y-intercept using \(x=0\).
• Find x-intercepts by solving \(ax^2+bx+c=0\).

If the parabola is given in conic form:

• Read the vertex directly.
• Use \(4p\) to find \(p\).
• Then find the focus and directrix using the formulas.

7. Common mistakes to avoid

• Sign errors: Be careful with negatives in \((x-h)\) and \((y-k)\). For example, \((x+3)^2\) means \(h=-3\).
• Mixing up axis and directrix: The axis of symmetry is a line through the vertex. The directrix is a different line used in the geometric definition.
• Forgetting to solve for y=0 when finding x-intercepts: x-intercepts happen where the graph crosses the x-axis.
• Confusing \(4p\) with \(p\): In conic form, the coefficient is \(4p\), so divide by 4 to get \(p\).

8. Why this matters

Understanding the anatomy of a parabola gives you more than just graphing skills. It helps you interpret maximum and minimum values, understand symmetry, and connect equations with geometric meaning.

These ideas also appear in real-world models such as projectile motion, satellite dishes, headlights, and optimization problems.

Brief Summary

A parabola has several key parts: the vertex, axis of symmetry, focus, directrix, and intercepts. From vertex form and standard form, you can find the turning point and intercepts. From conic form, you can also identify the focus and directrix. The more comfortable you become moving between these forms, the easier it becomes to analyze any quadratic graph.

Standard and Vertex Forms of a Quadratic

Quadratic functions are one of the most important types of functions in algebra. Their graphs are parabolas, and they appear in geometry, physics, optimization, and many real-world models.

In this lesson, you will learn how to work with two common ways of writing a quadratic function: standard form and vertex form. You will also learn how to convert between them by completing the square.

This skill is useful because each form tells you different information about the parabola. Standard form is convenient for algebraic operations, while vertex form makes the turning point of the graph easy to see.

1. The two forms of a quadratic

A quadratic in standard form is written as

where:

• \(a\), \(b\), and \(c\) are real numbers,
• \(a\neq 0\).

A quadratic in vertex form is written as

where the vertex of the parabola is the point \((h,k)\).

Why these forms matter:

• In standard form, the coefficients \(a\), \(b\), and \(c\) are easy to identify.
• In vertex form, the vertex is immediately visible.
• The value of \(a\) tells whether the parabola opens up or down, and how narrow or wide it is.

2. What the coefficient \(a\) tells you

• If \(a>0\), the parabola opens upward.
• If \(a<0\), the parabola opens downward.
• If \(|a|>1\), the parabola is narrower than \(y=x^2\).
• If \(0<|a|<1\), the parabola is wider than \(y=x^2\).

These ideas are true in both standard form and vertex form.

3. Understanding the vertex form

In vertex form,

the vertex is \((h,k)\). This is the highest or lowest point of the parabola.

• If \(a>0\), the vertex is a minimum point.
• If \(a<0\), the vertex is a maximum point.

The axis of symmetry is the vertical line

So vertex form is especially useful when you need to graph a parabola quickly or find its maximum or minimum value.

4. Understanding the standard form

In standard form,

the constant term \(c\) is the y-intercept, because when \(x=0\),

So the graph crosses the y-axis at \((0,c)\).

Standard form does not show the vertex right away, but it is still very useful. Many quadratics are first given in this form, so we often need to rewrite them into vertex form.

5. Converting from vertex form to standard form

To go from vertex form to standard form, expand the square and simplify.

Example:

First expand \((x-3)^2\):

Now multiply by 2:

So the standard form is

6. Converting from standard form to vertex form

To go from standard form to vertex form, we use completing the square.

This is the main skill in this lesson.

Start with

If \(a=1\), the process is simpler. If \(a\neq 1\), first factor \(a\) out of the \(x^2\) and \(x\) terms.

Completing the square pattern

Recall that

So if you have an expression like

you add and subtract

to create a perfect square trinomial.

That is,

and then

7. Worked Example 1: Simple conversion

Convert \(y=x^2+6x+2\) to vertex form.

Step 1: Group the \(x^2\) and \(x\) terms.

Step 2: Take half of 6 and square it.

Step 3: Add and subtract 9 inside the expression.

Step 4: Rewrite the perfect square trinomial.

Step 5: Simplify.

So the vertex form is

The vertex is \((-3,-7)\).

8. Worked Example 2: Leading coefficient not equal to 1

Convert \(y=2x^2+8x+1\) to vertex form.

Step 1: Factor out 2 from the first two terms.

Step 2: Complete the square inside the parentheses.

Half of 4 is 2, and \(2^2=4\).

Step 3: Rewrite the perfect square.

Step 4: Distribute the 2 and simplify.

So the vertex form is

The vertex is \((-2,-7)\).

Important note: When \(a\neq 1\), complete the square after factoring out \(a\) from the quadratic and linear terms. This is a very common place for mistakes.

9. Worked Example 3: Converting back to standard form

Convert \(y=-3(x-1)^2+4\) to standard form.

Step 1: Expand the square.

Step 2: Multiply by \(-3\).

Step 3: Simplify.

So the standard form is

The parabola opens downward because \(a=-3\).

10. Worked Example 4: Finding the vertex from standard form by completing the square

Write \(y=-x^2+4x-1\) in vertex form and identify the maximum value.

Step 1: Factor out \(-1\) from the \(x^2\) and \(x\) terms.

Step 2: Complete the square inside the parentheses.

Half of \(-4\) is \(-2\), and \((-2)^2=4\).

Step 3: Rewrite the perfect square.

Step 4: Simplify.

So the vertex form is

The vertex is \((2,3)\).

Because the parabola opens downward, the vertex gives the maximum value. The maximum value is

and it occurs when \(x=2\).

11. A step-by-step method for completing the square

1. Start with the quadratic in standard form.
2. If necessary, factor out the coefficient of \(x^2\) from the first two terms.
3. Find half of the coefficient of \(x\) inside the parentheses.
4. Square that number.
5. Add and subtract this square inside the expression.
6. Rewrite the trinomial as a squared binomial.
7. Simplify to get vertex form.

12. Common mistakes to avoid

• Forgetting to factor out \(a\) before completing the square when \(a\neq 1\).
• Using the wrong sign in the vertex. For example, in \((x+3)^2-7\), the vertex is \((-3,-7)\), not \((3,-7)\).
• Forgetting to simplify constants after completing the square.
• Expanding incorrectly when going from vertex form to standard form.

13. Why vertex form is useful in optimization

In many applications, a quadratic models a quantity that reaches a maximum or minimum. Since the vertex gives the turning point, vertex form is very helpful in optimization problems.

For example, if a profit function or height function is quadratic, the vertex tells you the greatest or least value and where it occurs.

14. Quick comparison of the two forms

• Standard form: \(y=ax^2+bx+c\)
• Vertex form: \(y=a(x-h)^2+k\)

From standard form, you can easily see:

• the coefficient \(a\), which affects shape and direction,
• the y-intercept \(c\).

From vertex form, you can easily see:

• the vertex \((h,k)\),
• the axis of symmetry \(x=h\),
• the maximum or minimum value.

15. Brief summary

A quadratic can be written in standard form, \(y=ax^2+bx+c\), or vertex form, \(y=a(x-h)^2+k\). Standard form is useful for seeing the y-intercept and working algebraically, while vertex form clearly shows the vertex and the maximum or minimum value.

To change from standard form to vertex form, use completing the square. To change from vertex form to standard form, expand and simplify. With practice, these two forms become powerful tools for understanding and graphing parabolas.

Lesson: The Discriminant and Nature of Roots

When solving a quadratic equation, one important question is: what kind of roots will the equation have? A quadratic may have two different real roots, one repeated real root, or two complex roots.

The discriminant is the part of the quadratic formula that helps us answer this question quickly, without fully solving the equation.

A quadratic equation in standard form is:

The quadratic formula is:

The expression under the square root,

is called the discriminant.

We usually represent the discriminant by the letter \(D\):

The value of \(D\) tells us the number and type of roots of the quadratic equation.

Why does this work? In the quadratic formula, the square root determines what kind of value we get:

• If the number inside the square root is positive, we can take its square root and get real values.
• If it is zero, the square root is zero, so both solutions become the same.
• If it is negative, the square root is not real, so the roots are complex.

Rules for the discriminant

• If \(D > 0\), the quadratic has two distinct real roots.
• If \(D = 0\), the quadratic has one repeated real root (also called a double root).
• If \(D < 0\), the quadratic has two complex conjugate roots.

We can go one step further when \(D > 0\):

• If \(D\) is a perfect square, the roots are usually rational.
• If \(D\) is not a perfect square, the roots are usually irrational.

This is very useful because it lets us predict the roots before doing all the algebra.

Connection to the graph of a parabola

A quadratic function \(y = ax^2 + bx + c\) graphs as a parabola. The roots of the equation \(ax^2 + bx + c = 0\) are the \(x\)-intercepts of the graph.

• If there are two distinct real roots, the parabola crosses the \(x\)-axis at two points.
• If there is one repeated real root, the parabola touches the \(x\)-axis at exactly one point.
• If there are two complex roots, the parabola does not cross the \(x\)-axis.

So the discriminant also tells us how the parabola behaves relative to the \(x\)-axis.

How to use the discriminant

1. Write the quadratic in standard form: \(ax^2 + bx + c = 0\).
2. Identify \(a\), \(b\), and \(c\).
3. Calculate \(D = b^2 - 4ac\).
4. Use the sign of \(D\) to determine the nature of the roots.

Worked Example 1: Two distinct real roots

Determine the nature of the roots of:

Here, \(a = 1\), \(b = -5\), and \(c = 6\).

Find the discriminant:

Since \(D = 1\), which is positive, the equation has two distinct real roots.

Also, \(1\) is a perfect square, so the roots are rational.

In fact, if we solve the equation, we get:

This matches our prediction.

Worked Example 2: One repeated real root

Determine the nature of the roots of:

Here, \(a = 2\), \(b = 4\), and \(c = 2\).

Calculate the discriminant:

Since \(D = 0\), the equation has one repeated real root.

Using the quadratic formula:

So the root is:

This root occurs twice, so we can also write the factorized form as:

Worked Example 3: Two complex roots

Determine the nature of the roots of:

Here, \(a = 1\), \(b = 2\), and \(c = 5\).

Find the discriminant:

Since \(D = -16\), which is negative, the equation has two complex conjugate roots.

Now solve to see them:

Because \(\sqrt{-16} = 4i\), we get:

So the roots are:

The parabola does not cross the \(x\)-axis because there are no real roots.

Worked Example 4: Using the discriminant without solving fully

Determine the nature of the roots of:

Identify the coefficients:

Compute the discriminant:

Since \(73 > 0\), the equation has two distinct real roots.

Since \(73\) is not a perfect square, the roots are irrational.

Notice that we were able to describe the roots without finding their exact values.

Special note about complex conjugates

If a quadratic equation has real coefficients and its discriminant is negative, the roots come in a pair called complex conjugates.

For example, if one root is \(3 + 2i\), the other must be \(3 - 2i\).

This happens because the coefficients \(a\), \(b\), and \(c\) are real numbers.

Common mistakes to avoid

• Forgetting standard form: Make sure the equation is written as \(ax^2 + bx + c = 0\) before identifying \(a\), \(b\), and \(c\).
• Sign errors: Be careful with negative values of \(b\) and \(c\). For example, if \(b = -5\), then \(b^2 = 25\), not \(-25\).
• Mixing up the cases: Positive discriminant means two real roots, zero means one repeated root, negative means complex roots.
• Confusing one root with one repeated root: When \(D = 0\), there are technically two equal roots, not just a single unrelated answer.

Quick reference table

• \(D > 0\): two distinct real roots
• \(D = 0\): one repeated real root
• \(D < 0\): two complex conjugate roots

Why this matters

The discriminant is useful in algebra because it tells you what to expect before solving.

It is also useful in graphing, since it shows how many times a parabola meets the \(x\)-axis.

As you continue studying quadratic functions and complex numbers, the discriminant becomes an important link between equations, graphs, and the number system.

Brief Summary

The discriminant of a quadratic equation is \(D = b^2 - 4ac\). It comes from the quadratic formula and tells us the nature of the roots. If \(D > 0\), there are two distinct real roots. If \(D = 0\), there is one repeated real root. If \(D < 0\), there are two complex conjugate roots. By checking the discriminant, we can predict the roots and understand the graph of the parabola more easily.

Lesson: The Imaginary Unit and Complex Arithmetic

In earlier algebra, you learned that some equations have no real solutions. For example, the equation \(x^2=-1\) has no real number solution, because the square of any real number is never negative.

To solve equations like this, mathematics extends the real number system by introducing a new number called the imaginary unit, written as \(i\). This allows us to work with complex numbers, which combine real and imaginary parts.

Complex numbers are important because they make algebra more complete. In fact, using complex numbers means that polynomial equations, including quadratics, can have roots even when those roots are not real.

1. The imaginary unit

The imaginary unit is defined by the rule

This definition is the foundation of all complex arithmetic. Whenever you see \(i^2\), you can replace it with \(-1\).

For example:

• \(i^2=-1\)
• \(3i\) means 3 times \(i\)
• \(\sqrt{-9}=3i\)
• \(\sqrt{-25}=5i\)

Be careful: in school algebra, we usually write square roots of negative numbers in terms of \(i\), but we do not treat square roots exactly the same way as positive roots in every situation. For now, focus on simplifying expressions correctly using \(i^2=-1\).

2. Standard form of a complex number

A complex number is any number that can be written in the form

where \(a\) and \(b\) are real numbers.

• \(a\) is called the real part
• \(b\) is called the imaginary part

Examples of complex numbers in standard form:

• \(4+3i\)
• \(-2-5i\)
• \(7+0i\), which is just the real number 7
• \(0+6i\), which is a purely imaginary number

3. Powers of \(i\)

Higher powers of \(i\) follow a repeating pattern. Start by multiplying step by step:

Then the pattern repeats every 4 powers:

This means:

• If the exponent is a multiple of 4, then \(i^n=1\)
• If the remainder when dividing by 4 is 1, then \(i^n=i\)
• If the remainder is 2, then \(i^n=-1\)
• If the remainder is 3, then \(i^n=-i\)

For example:

• \(i^6=i^{4+2}=i^4\cdot i^2=1\cdot(-1)=-1\)
• \(i^{11}=i^{8+3}=i^8\cdot i^3=1\cdot(-i)=-i\)
• \(i^{20}=1\) because 20 is a multiple of 4

4. Adding and subtracting complex numbers

To add or subtract complex numbers, combine like terms:

• real parts with real parts
• imaginary parts with imaginary parts

Example structure:

This is very similar to combining like terms in algebra.

Worked Example 1: Addition and subtraction

Simplify:

Combine real parts and imaginary parts:

Now simplify:

So the answers are \(5-2i\) and \(5-7i\).

5. Multiplying complex numbers

To multiply complex numbers, use the distributive property, also called FOIL when multiplying two binomials. Then simplify using \(i^2=-1\).

General form:

Since \(i^2=-1\), this becomes

Worked Example 2: Multiplication

Simplify:

Use distribution:

Now use \(i^2=-1\):

So,

6. Dividing complex numbers

Division is the most challenging operation with complex numbers. The goal is to write the final answer in standard form \(a+bi\). To do this, we remove \(i\) from the denominator.

We use the complex conjugate. The conjugate of

is

The conjugate has the same real part but the opposite imaginary part.

This is useful because:

Notice that the result is real, because \(i^2=-1\).

So when dividing by a complex number, multiply both numerator and denominator by the conjugate of the denominator.

Worked Example 3: Division

Simplify:

The conjugate of the denominator \(1-2i\) is \(1+2i\). Multiply top and bottom by \(1+2i\):

First simplify the numerator:

Now simplify the denominator:

So the fraction becomes

Therefore,

7. Simplifying expressions with powers of \(i\)

When simplifying powers of \(i\), reduce the exponent using the repeating cycle of 4.

Worked Example 4: Higher powers and mixed operations

Simplify:

Reduce each power:

• \(15 \div 4\) leaves remainder 3, so \(i^{15}=i^3=-i\)
• \(6 \div 4\) leaves remainder 2, so \(i^6=i^2=-1\)
• \(9 \div 4\) leaves remainder 1, so \(i^9=i\)

Substitute these values:

So the simplified result is

8. Common mistakes to avoid

• Forgetting that \(i^2=-1\): This is the most important rule in complex arithmetic.
• Not writing answers in standard form: Final answers should usually look like \(a+bi\).
• Combining unlike terms: Real and imaginary parts should stay separate unless you are simplifying \(i^2\).
• Stopping too early in multiplication: Always simplify any \(i^2\) terms.
• Dividing without using the conjugate: If there is an imaginary part in the denominator, multiply by the conjugate.

9. Why this matters for quadratic equations

Complex numbers connect directly to quadratic functions and equations. For example, consider:

Solving gives

These are not real roots, but they are valid complex roots. This shows how complex numbers allow quadratics to have solutions even when the graph does not cross the \(x\)-axis.

10. Key rules to remember

• \(i=\sqrt{-1}\)
• \(i^2=-1\)
• Complex numbers are written as \(a+bi\)
• Add and subtract by combining like terms
• Multiply using distribution, then replace \(i^2\) with \(-1\)
• Divide by multiplying by the conjugate of the denominator
• Powers of \(i\) repeat every 4 terms

Brief Summary

The imaginary unit \(i\) is defined so that \(i^2=-1\), which allows us to work with square roots of negative numbers. A complex number has the form \(a+bi\), where \(a\) is the real part and \(b\) is the imaginary part.

You can add and subtract complex numbers by combining like terms, multiply them using distribution and \(i^2=-1\), and divide them by using the conjugate. Once you understand these rules, you can simplify powers of \(i\) and solve quadratic equations that have non-real roots.

Complex Conjugates and Division is an important idea in complex numbers. It helps us divide complex expressions and rewrite answers in standard form, which is the form \(a+bi\), where \(a\) and \(b\) are real numbers.

This skill is especially useful when a complex number appears in the denominator of a fraction. Just like we rationalize denominators involving square roots, we can also simplify denominators involving imaginary numbers by using a complex conjugate.

In this lesson, you will learn what a complex conjugate is, why it is useful, and how to use it to divide complex numbers step by step.

1. Review: What is a complex number?

A complex number has the form \(a+bi\), where:

• \(a\) is the real part
• \(b\) is the imaginary part
• \(i\) is the imaginary unit, and \(i^2=-1\)

Examples of complex numbers include:

• \(3+2i\)
• \(-5-i\)
• \(4i\), which can be written as \(0+4i\)
• \(7\), which can be written as \(7+0i\)

2. What is a complex conjugate?

The complex conjugate of a complex number changes the sign of the imaginary part.

If a complex number is \(a+bi\), then its conjugate is:

$$a-bi$$

If a complex number is \(a-bi\), then its conjugate is:

$$a+bi$$

Examples:

• The conjugate of \(3+4i\) is \(3-4i\)
• The conjugate of \(7-2i\) is \(7+2i\)
• The conjugate of \(-5+i\) is \(-5-i\)

3. Why are conjugates useful?

When we multiply a complex number by its conjugate, the imaginary terms cancel out. The result is a real number.

For example:

$$ (3+4i)(3-4i) $$

Use the distributive property:

$$ =9-12i+12i-16i^2 $$

The middle terms cancel:

$$ =9-16(-1) $$

Since \(i^2=-1\):

$$ =9+16=25 $$

This always happens in the same pattern:

$$ (a+bi)(a-bi)=a^2+b^2 $$

Notice that the result is real because:

• the \(bi\) terms cancel
• \(i^2=-1\), which turns the last term into a positive real number

This is exactly why conjugates help with division. They turn a complex denominator into a real number.

4. Dividing complex numbers

Suppose you want to simplify:

$$ \frac{a+bi}{c+di} $$

If the denominator contains \(i\), we do not leave the answer that way. We rewrite the result in standard form \(x+yi\).

To do this:

1. Take the conjugate of the denominator.
2. Multiply both the numerator and denominator by that conjugate.
3. Simplify the denominator to a real number.
4. Write the final answer in the form \(a+bi\).

For example, if the denominator is \(c+di\), multiply by:

$$ \frac{c-di}{c-di} $$

This is equal to 1, so the value of the expression does not change.

Then the denominator becomes:

$$ (c+di)(c-di)=c^2+d^2 $$

That gives a real denominator, which is much easier to work with.

5. Worked Example 1: Basic division by an imaginary number

Simplify:

$$ \frac{6}{i} $$

The denominator is \(i\). Its conjugate is \(-i\).

Multiply numerator and denominator by \(-i\):

$$ \frac{6}{i}\cdot\frac{-i}{-i}=\frac{-6i}{-i^2} $$

Since \(i^2=-1\):

$$ \frac{-6i}{-(-1)}=\frac{-6i}{1}=-6i $$

Answer: \(-6i\)

This shows that even a denominator with just \(i\) can be simplified using the same idea.

6. Worked Example 2: Dividing a real number by a complex number

Simplify:

$$ \frac{5}{2-3i} $$

The conjugate of \(2-3i\) is \(2+3i\).

Multiply top and bottom by \(2+3i\):

$$ \frac{5}{2-3i}\cdot\frac{2+3i}{2+3i} =\frac{5(2+3i)}{(2-3i)(2+3i)} $$

Simplify the numerator:

$$ 5(2+3i)=10+15i $$

Simplify the denominator using conjugates:

$$ (2-3i)(2+3i)=2^2+3^2=4+9=13 $$

So:

$$ \frac{10+15i}{13}=\frac{10}{13}+\frac{15}{13}i $$

Answer: \(\frac{10}{13}+\frac{15}{13}i\)

7. Worked Example 3: Dividing one complex number by another

Simplify:

$$ \frac{3+2i}{1-4i} $$

The conjugate of the denominator \(1-4i\) is \(1+4i\).

Multiply numerator and denominator by \(1+4i\):

$$ \frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i} =\frac{(3+2i)(1+4i)}{(1-4i)(1+4i)} $$

Now expand the numerator:

$$ (3+2i)(1+4i)=3+12i+2i+8i^2 $$

Combine like terms and use \(i^2=-1\):

$$ =3+14i+8(-1) $$

$$ =3+14i-8=-5+14i $$

Now simplify the denominator:

$$ (1-4i)(1+4i)=1^2+4^2=1+16=17 $$

So the fraction becomes:

$$ \frac{-5+14i}{17}=-\frac{5}{17}+\frac{14}{17}i $$

Answer: \(-\frac{5}{17}+\frac{14}{17}i\)

8. Worked Example 4: Division with more algebra

Simplify:

$$ \frac{4-7i}{2+i} $$

The conjugate of \(2+i\) is \(2-i\).

Multiply numerator and denominator by \(2-i\):

$$ \frac{4-7i}{2+i}\cdot\frac{2-i}{2-i} =\frac{(4-7i)(2-i)}{(2+i)(2-i)} $$

Expand the numerator:

$$ (4-7i)(2-i)=8-4i-14i+7i^2 $$

Since \(i^2=-1\):

$$ =8-18i+7(-1) $$

$$ =8-18i-7=1-18i $$

Now simplify the denominator:

$$ (2+i)(2-i)=2^2+1^2=4+1=5 $$

So:

$$ \frac{1-18i}{5}=\frac{1}{5}-\frac{18}{5}i $$

Answer: \(\frac{1}{5}-\frac{18}{5}i\)

9. A useful shortcut formula

For a quotient like

$$ \frac{a+bi}{c+di} $$

multiply by the conjugate \(c-di\):

$$ \frac{a+bi}{c+di}\cdot\frac{c-di}{c-di} $$

After simplifying, the denominator becomes:

$$ c^2+d^2 $$

The numerator can be expanded carefully using distribution. Although some students memorize a formula, it is usually safer in Grade 12 to expand step by step and simplify. That helps avoid sign mistakes.

10. Common mistakes to avoid

• Forgetting to multiply both numerator and denominator
  You must multiply the entire fraction by the conjugate over itself.
• Using the wrong conjugate
  Only the sign of the imaginary part changes. For example, the conjugate of \(5-2i\) is \(5+2i\), not \(-5+2i\).
• Forgetting that \(i^2=-1\)
  This is one of the most common errors. For example, \(6i^2=-6\), not \(6\).
• Not writing the final answer in standard form
  Your final answer should look like \(a+bi\).
• Sign errors when expanding
  Be careful with negative signs, especially when multiplying terms like \((-3i)(4i)\).

11. Quick check: Why does the denominator become real?

If the denominator is \(a+bi\), multiplying by its conjugate gives:

$$ (a+bi)(a-bi)=a^2-b^2i^2 $$

Since \(i^2=-1\):

$$ =a^2-b^2(-1)=a^2+b^2 $$

This is a real number. That is the key reason conjugates are used in division.

12. When should you use conjugates?

Use complex conjugates when:

• you are dividing by a complex number
• you need to remove \(i\) from the denominator
• you are asked to write the answer in standard form \(a+bi\)

Summary

A complex conjugate is formed by changing the sign of the imaginary part of a complex number. Conjugates are useful because multiplying a complex number by its conjugate gives a real number. To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator, simplify, and write the result in the form \(a+bi\). With careful expansion and correct use of \(i^2=-1\), you can simplify any complex division problem.

Roots Over the Complex Plane helps us solve quadratic equations even when the graph does not cross the x-axis.

In earlier algebra, you may have learned that some quadratic equations have “no real solutions.” For example, the equation \(x^2+1=0\) cannot be solved using real numbers, because no real number squared gives \(-1\).

To solve equations like this, we extend the real number system to the complex number system. This allows every quadratic equation to have roots, which is an important idea called algebraic closure.

In this lesson, you will learn how to:

• recognize when a quadratic has complex roots,
• calculate those roots using the quadratic formula,
• write answers in standard complex form, and
• understand what these roots mean graphically.

1. Review: The quadratic formula

For any quadratic equation

the solutions are given by

The expression under the square root, \(b^2-4ac\), is called the discriminant.

The discriminant tells us what kind of roots the quadratic has:

• If \(b^2-4ac>0\), there are two distinct real roots.
• If \(b^2-4ac=0\), there is one repeated real root.
• If \(b^2-4ac<0\), there are two complex roots.

Our focus is the third case: when the discriminant is negative.

2. The imaginary unit

To work with square roots of negative numbers, we use the imaginary unit \(i\), defined by

This means

Using \(i\), we can rewrite square roots of negative numbers:

• \(\sqrt{-4}=\sqrt{4}\sqrt{-1}=2i\)
• \(\sqrt{-9}=3i\)
• \(\sqrt{-20}=\sqrt{20}i=2\sqrt{5}i\)

A complex number is written in the form

where \(a\) is the real part and \(b\) is the imaginary part.

3. Complex roots from a negative discriminant

Suppose the discriminant is negative:

Then \(\sqrt{b^2-4ac}\) involves the square root of a negative number. We rewrite it using \(i\).

For example, if

then

So the quadratic formula becomes

This produces two complex solutions.

4. Complex roots come in conjugate pairs

If a quadratic equation has real coefficients and one root is complex, then the other root must be its complex conjugate.

If one root is

then the other is

These two numbers are mirror images across the real axis on the complex plane.

This is why quadratics with negative discriminants usually have answers that look like

5. The complex plane

Complex numbers can be shown on a graph called the complex plane.

• The horizontal axis is the real axis.
• The vertical axis is the imaginary axis.

For example, the complex number

is graphed at the point \((3,2)\).

The complex number

is graphed at the point \((3,-2)\).

So when a quadratic has roots \(3\pm 2i\), the roots are two points vertically opposite each other across the real axis.

6. Graphical meaning for quadratics

A quadratic function

is a parabola on the usual coordinate plane.

The roots of the quadratic are the x-values where

These are the x-intercepts of the graph.

If the discriminant is negative, the parabola does not cross the x-axis, so there are no real x-intercepts.

However, the equation still has solutions in the complex number system. These roots are not visible as x-intercepts on the usual graph, but they can be represented as points on the complex plane.

This shows an important idea: a quadratic may have no real roots, but it still has roots over the complex plane.

7. Step-by-step method

When solving a quadratic with possible complex roots, use this process:

1. Write the equation in standard form: \(ax^2+bx+c=0\).
2. Identify \(a\), \(b\), and \(c\).
3. Find the discriminant: \(b^2-4ac\).
4. If the discriminant is negative, rewrite its square root using \(i\).
5. Simplify the result.
6. Write the final answers in the form \(a+bi\) and \(a-bi\) if possible.

Worked Example 1: A simple quadratic with pure imaginary roots

Solve

Step 1: Move to isolate \(x^2\):

Step 2: Take square roots:

Step 3: Rewrite using \(i\):

Answer:

Graphical meaning: The parabola \(y=x^2+9\) lies entirely above the x-axis, so it has no real x-intercepts. On the complex plane, the roots are the points \((0,3)\) and \((0,-3)\).

Worked Example 2: Using the quadratic formula

Solve

Step 1: Identify values:

Step 2: Find the discriminant:

The discriminant is negative, so the roots are complex.

Step 3: Apply the quadratic formula:

Step 4: Rewrite the square root:

Step 5: Simplify:

Answer:

Graphical meaning: The parabola \(y=x^2+4x+8\) does not meet the x-axis. On the complex plane, the roots are \((-2,2)\) and \((-2,-2)\).

Worked Example 3: Simplifying a non-perfect square negative discriminant

Solve

Step 1: Identify values:

Step 2: Find the discriminant:

Step 3: Use the quadratic formula:

Step 4: Rewrite \(\sqrt{-40}\):

So

Step 5: Simplify:

Answer:

Notice that the real part is the same in both roots, and the imaginary parts are opposites.

Worked Example 4: Connecting roots, discriminant, and graph

Consider the quadratic

Step 1: To find the roots, solve

Here,

Step 2: Find the discriminant:

Step 3: Use the quadratic formula:

Since \(\sqrt{-16}=4i\),

Answer:

Graphical interpretation:

• On the usual graph of the parabola, there are no x-intercepts.
• The vertex is at \(x=3\), and the parabola opens upward.
• On the complex plane, the roots are the points \((3,2)\) and \((3,-2)\).

This example shows that the real part of the roots often matches the axis of symmetry of the parabola.

8. Common mistakes to avoid

• Forgetting the \(\pm\): Square roots in quadratic equations usually give two answers.
• Leaving \(\sqrt{-n}\) unsimplified: Rewrite it as \(i\sqrt{n}\).
• Confusing \(-\sqrt{n}\) with \(\sqrt{-n}\): These are not the same. For example, \(-\sqrt{9}=-3\), but \(\sqrt{-9}=3i\).
• Not simplifying fully: Reduce fractions and simplify radicals when possible.
• Thinking “no real roots” means “no roots”: Over the complex numbers, the quadratic still has solutions.

9. Quick check ideas

You can quickly predict complex roots by checking the discriminant.

For example, for

the discriminant is

Since it is negative, the roots must be complex.

This can save time before you begin solving.

10. Why this matters

Complex roots complete the picture of quadratic equations. Without complex numbers, some equations would seem unsolvable. With complex numbers, every quadratic has roots.

This idea is useful in higher algebra, science, engineering, and many areas where equations describe patterns, motion, and change.

Brief Summary

When the discriminant \(b^2-4ac\) of a quadratic is negative, the roots are complex. We use the imaginary unit \(i\), where \(i^2=-1\), to rewrite square roots of negative numbers.

Complex roots are written in the form \(a+bi\) and usually appear as conjugate pairs \(a\pm bi\). A quadratic with complex roots has no real x-intercepts on its parabola, but its roots can still be shown as points on the complex plane.

Vieta’s Formulas for Quadratics are a powerful shortcut for connecting the roots of a quadratic equation to its coefficients. Instead of solving the equation first, we can often find information about the roots directly from the equation itself.

This is especially useful when you need to:

• find the sum or product of the roots quickly,
• build a quadratic equation from given roots,
• check whether roots make sense,
• work with real or complex roots without solving completely.

In this lesson, you will learn what Vieta’s formulas are, where they come from, and how to use them in different types of problems.

1. The standard quadratic form

A quadratic equation is usually written as

$$ax^2+bx+c=0$$

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

Suppose the two roots of this equation are \(r_1\) and \(r_2\). These roots may be real or complex. Then the quadratic can also be written in factored form as

$$a(x-r_1)(x-r_2)=0$$

Now expand the factors:

$$a(x-r_1)(x-r_2)=a\left(x^2-(r_1+r_2)x+r_1r_2\right)$$

$$=ax^2-a(r_1+r_2)x+a(r_1r_2)$$

Compare this with the standard form

$$ax^2+bx+c$$

Matching the coefficients gives:

$$-a(r_1+r_2)=b$$

$$a(r_1r_2)=c$$

Now divide by \(a\):

$$r_1+r_2=-\frac{b}{a}$$

$$r_1r_2=\frac{c}{a}$$

These are Vieta’s formulas for quadratics.

2. The formulas to remember

For the quadratic equation

$$ax^2+bx+c=0$$

with roots \(r_1\) and \(r_2\),

• Sum of the roots: \(r_1+r_2=-\frac{b}{a}\)
• Product of the roots: \(r_1r_2=\frac{c}{a}\)

A common mistake is forgetting the negative sign in the sum formula. The sum is negative \(b/a\), not just \(b/a\).

3. Why these formulas matter

Vieta’s formulas help you understand a quadratic even before you solve it. For example, if the sum of the roots is positive and the product is positive, that tells you something about how the roots are related.

They also work whether the roots are:

• two different real numbers,
• one repeated real number,
• or two complex conjugates.

So Vieta’s formulas are part of the bigger idea that every quadratic has roots in the complex number system, even if they are not real.

4. Using Vieta’s formulas directly

If you are given a quadratic equation, you do not need to solve it to find the sum and product of the roots. Just identify \(a\), \(b\), and \(c\), then apply the formulas.

Worked Example 1: Find the sum and product of the roots

Find the sum and product of the roots of

$$2x^2-7x+3=0$$

Here, \(a=2\), \(b=-7\), and \(c=3\).

Use Vieta’s formulas:

$$r_1+r_2=-\frac{b}{a}=-\frac{-7}{2}=\frac{7}{2}$$

$$r_1r_2=\frac{c}{a}=\frac{3}{2}$$

Answer:

• Sum of the roots = \(\frac{7}{2}\)
• Product of the roots = \(\frac{3}{2}\)

Notice that we found both values without solving the equation.

5. Checking by solving

Sometimes it helps to verify that Vieta’s formulas agree with the actual roots.

Worked Example 2: Solve and check

Consider

$$x^2-5x+6=0$$

First use Vieta’s formulas:

$$r_1+r_2=-\frac{-5}{1}=5$$

$$r_1r_2=\frac{6}{1}=6$$

Now solve the equation by factoring:

$$x^2-5x+6=(x-2)(x-3)=0$$

So the roots are \(2\) and \(3\).

Check:

$$2+3=5$$

$$2\cdot 3=6$$

The results match Vieta’s formulas perfectly.

6. Forming a quadratic equation from given roots

Vieta’s formulas also work in reverse. If you know the roots, you can build the quadratic equation.

If the roots are \(r_1\) and \(r_2\), then a quadratic with those roots is

$$x^2-(r_1+r_2)x+r_1r_2=0$$

This comes from

$$ (x-r_1)(x-r_2)=0 $$

when the leading coefficient is 1.

Worked Example 3: Build the equation from roots

Find a quadratic equation whose roots are \(4\) and \(-1\).

First find the sum and product:

$$r_1+r_2=4+(-1)=3$$

$$r_1r_2=4(-1)=-4$$

Now substitute into the form

$$x^2-(r_1+r_2)x+r_1r_2=0$$

$$x^2-3x-4=0$$

Answer: A quadratic equation with roots \(4\) and \(-1\) is

$$x^2-3x-4=0$$

You can check by factoring:

$$x^2-3x-4=(x-4)(x+1)$$

7. Using Vieta’s formulas with complex roots

Quadratics can have complex roots. Vieta’s formulas still work exactly the same way.

For quadratics with real coefficients, non-real roots come in conjugate pairs. That means if one root is \(a+bi\), the other is \(a-bi\).

Worked Example 4: Complex roots

Consider the equation

$$x^2-6x+13=0$$

Using Vieta’s formulas:

$$r_1+r_2=6$$

$$r_1r_2=13$$

Now solve to see the roots:

$$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(13)}}{2(1)}$$

$$x=\frac{6\pm\sqrt{36-52}}{2}$$

$$x=\frac{6\pm\sqrt{-16}}{2}$$

$$x=\frac{6\pm 4i}{2}=3\pm 2i$$

So the roots are \(3+2i\) and \(3-2i\).

Check the sum:

$$ (3+2i)+(3-2i)=6 $$

Check the product:

$$ (3+2i)(3-2i)=9-(2i)^2=9+4=13 $$

Again, Vieta’s formulas work perfectly.

8. Special cases and observations

• If the product of the roots is negative, the roots have opposite signs when they are real.
• If the product is positive and the sum is negative, both real roots are often negative.
• If the discriminant is zero, the two roots are equal, and Vieta’s formulas still hold.

For example, in

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

the roots are both \(2\). Then:

$$2+2=4=-\frac{-4}{1}$$

$$2\cdot 2=4=\frac{4}{1}$$

9. Common mistakes to avoid

• Forgetting the negative sign in \(r_1+r_2=-\frac{b}{a}\).
• Mixing up sum and product. The sum uses \(b\); the product uses \(c\).
• Ignoring the leading coefficient \(a\). If \(a\neq 1\), you must divide by \(a\).
• Assuming roots must be real. Vieta’s formulas also apply to complex roots.

10. Quick strategy for problems

1. Write the quadratic in the form \(ax^2+bx+c=0\).
2. Identify \(a\), \(b\), and \(c\) carefully.
3. Use:

$$r_1+r_2=-\frac{b}{a}, \qquad r_1r_2=\frac{c}{a}$$

4. If needed, use these values to form a new quadratic or check roots.

11. Brief summary

Vieta’s formulas connect the roots of a quadratic equation directly to its coefficients. For

$$ax^2+bx+c=0$$

with roots \(r_1\) and \(r_2\),

$$r_1+r_2=-\frac{b}{a}, \qquad r_1r_2=\frac{c}{a}$$

These formulas let you find the sum and product of roots without solving, create a quadratic from known roots, and work with both real and complex solutions. They are one of the most useful tools for understanding quadratic equations efficiently.

Quadratic Inequalities and Sign Charts

When you solve a quadratic equation, you are looking for the values of \(x\) that make the expression equal to 0.

When you solve a quadratic inequality, you are looking for the values of \(x\) that make the expression positive or negative.

For example, instead of solving \(x^2-5x+6=0\), you might solve:

or

To solve these, we use a very useful tool called a sign chart.

A sign chart helps us track where a quadratic expression is:

• positive \((+)\)
• negative \((- )\)
• zero

This lesson will show you how to find those intervals and write the solution correctly.

1. What is a quadratic inequality?

A quadratic inequality is an inequality that involves a quadratic expression, such as:

• \(ax^2+bx+c>0\)
• \(ax^2+bx+c<0\)
• \(ax^2+bx+c\ge 0\)
• \(ax^2+bx+c\le 0\)

Since a quadratic expression graphs as a parabola, solving the inequality means identifying where the parabola is above or below the \(x\)-axis.

2. The key idea behind sign charts

A quadratic can only change sign at values where it equals 0. These values are called its critical values or zeros.

So the first step is always to solve:

These zero values split the number line into intervals. On each interval, the sign of the quadratic stays the same.

That means we can:

1. Find the zeros.
2. Mark them on a number line.
3. Test one point from each interval.
4. Determine whether the expression is positive or negative on that interval.

This is the idea of a sign chart.

3. General method for solving quadratic inequalities

1. Move everything to one side so one side is 0.
2. Factor the quadratic if possible, or find its roots another way.
3. Find the critical values by setting the quadratic equal to 0.
4. Divide the number line into intervals using those values.
5. Test a point in each interval.
6. Choose the intervals that satisfy the inequality.
7. Include or exclude endpoints depending on the symbol:

• Use parentheses for \(<\) or \(>\), because the roots are not included.
• Use brackets for \(\le\) or \(\ge\), because the roots are included if they make the expression 0.

4. Understanding the sign of a quadratic

If a quadratic factors as

then the roots are \(r_1\) and \(r_2\). These roots split the number line into up to three intervals:

Inside each interval, the sign is constant.

Also remember the shape of the parabola:

• If \(a>0\), the parabola opens upward.
• If \(a<0\), the parabola opens downward.

This helps you predict whether the quadratic is positive outside the roots or between the roots.

For a quadratic with two distinct real roots:

• If \(a>0\), the expression is positive outside the roots and negative between them.
• If \(a<0\), the expression is negative outside the roots and positive between them.

5. Worked Example 1

Solve:

Step 1: Factor the quadratic

So we solve the related equation:

The critical values are:

Step 2: Set up intervals

The roots divide the number line into:

Step 3: Test one point in each interval

• For \(( -\infty,2)\), test \(x=0\):

• For \((2,3)\), test \(x=2.5\):

• For \((3,\infty)\), test \(x=4\):

Step 4: Choose the intervals where the expression is positive

We want \(>0\), so the solution is:

Sign chart:

\(+\) on \(( -\infty,2)\), \(-\) on \((2,3)\), \(+\) on \((3,\infty)\).

6. Worked Example 2

Solve:

This is the same quadratic as before, so the factorization is:

From the previous sign chart:

• positive on \(( -\infty,2)\)
• negative on \((2,3)\)
• positive on \((3,\infty)\)

We now want values where the expression is negative or zero.

It is negative between the roots and equal to zero at the roots.

So the solution is:

Notice the square brackets. They are used because \(x=2\) and \(x=3\) make the expression equal to 0, and \(\le 0\) includes zero.

7. Worked Example 3

Solve:

Step 1: Find the roots

Factor the quadratic:

Now factor the inside:

So:

The roots are found from:

So:

The critical values are \(-1\) and \(\frac{3}{2}\).

Step 2: Set up intervals

Step 3: Test points

• Test \(x=-2\):

• Test \(x=0\):

• Test \(x=2\):

So the sign chart is:

\(-\) on \(( -\infty,-1)\), \(+\) on \((-1,\tfrac{3}{2})\), \(-\) on \((\tfrac{3}{2},\infty)\).

Step 4: Choose where the expression is nonnegative

We want \(\ge 0\), so we include the positive interval and the roots.

This example is important because the leading coefficient is negative, so the signs are reversed from the usual upward-opening case.

8. Worked Example 4

Solve:

Step 1: Try to find real roots

Set the quadratic equal to 0:

Use the discriminant:

Since the discriminant is negative, the quadratic has no real roots.

That means the expression never crosses the \(x\)-axis, so its sign never changes.

Step 2: Determine whether it is always positive or always negative

Test any value, such as \(x=0\):

So the expression is positive for all real \(x\).

The solution is:

This example connects to complex numbers: the equation has no real roots, only complex roots, so there are no real critical values to place on the sign chart. Because of that, the quadratic keeps the same sign for all real \(x\).

9. Special case: a repeated root

Sometimes a quadratic has one repeated real root, such as:

This equals 0 only at \(x=2\).

Because the factor is squared, the sign does not change when crossing \(x=2\).

For example, \((x-2)^2\ge 0\) for every real number, since a square is never negative.

So if a quadratic has a repeated root, be careful: the sign chart may touch 0 and stay with the same sign on both sides.

10. Common mistakes to avoid

• Forgetting to move all terms to one side. You want the inequality in the form \(ax^2+bx+c \; ? \; 0\).
• Using the roots but not testing intervals. The roots alone do not tell you the answer unless you know the sign in each interval.
• Including endpoints when you should not. For \(>\) or \(<\), do not include the roots.
• Ignoring the leading coefficient. If the quadratic opens downward, the positive and negative regions switch compared with an upward-opening parabola.
• Assuming every quadratic has two real roots. Some have one repeated root, and some have no real roots.

11. Quick strategy guide

When solving a quadratic inequality, ask yourself these questions:

1. Is everything on one side and 0 on the other?
2. Can I factor the quadratic?
3. What are the real roots?
4. What intervals do those roots create?
5. Is the expression positive or negative on each interval?
6. Does the inequality include equality?

12. Summary

Quadratic inequalities are solved by finding where a quadratic expression is positive, negative, or zero.

A sign chart is built from the real zeros of the quadratic. These zeros split the number line into intervals, and testing one point in each interval tells you the sign of the expression there.

For an upward-opening quadratic with two real roots, the expression is usually positive outside the roots and negative between them. For a downward-opening quadratic, it is the opposite.

If there are no real roots, the sign never changes on the real number line. If there is a repeated root, the graph touches the axis but may not cross it.

With practice, sign charts become a quick and reliable way to solve quadratic inequalities.

Quadratic Optimization Models are used when a real-world quantity can be described by a quadratic function and we want to find its maximum or minimum value.

In Grade 12 Maths, this often appears in problems about area, revenue, and projectile motion. These situations are important because many real-world systems increase for a while, reach a highest or lowest point, and then change direction. That pattern is often modeled by a parabola.

This lesson will show you how to build a quadratic model from a situation, identify what quantity is being optimized, and use the vertex of the parabola to find the maximum or minimum value.

1. What is a quadratic model?

A quadratic function has the form

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

where \(a\neq 0\).

When graphed, a quadratic function forms a parabola.

• If \(a>0\), the parabola opens upward, so it has a minimum value.
• If \(a<0\), the parabola opens downward, so it has a maximum value.

The key point in optimization is the vertex, because it gives the highest or lowest value of the function.

2. The vertex and optimization

For a quadratic function

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

the x-coordinate of the vertex is

$$x=-\frac{b}{2a}$$

Then substitute that value into the function to get the maximum or minimum value.

So the basic optimization process is:

1. Define the variable.
2. Write an expression for the quantity to optimize.
3. Simplify it into quadratic form.
4. Use \(x=-\frac{b}{2a}\) to find the vertex.
5. Interpret the result in the context of the problem.

3. Why quadratic optimization works

Many applied problems involve one quantity depending on another in a curved way rather than a straight-line way. For example, increasing the width of a rectangle may increase area at first, but after a certain point the length becomes too small and area starts decreasing.

This rise-then-fall pattern creates a parabola that has a turning point. That turning point is exactly what optimization is looking for.

4. Common types of quadratic optimization problems

• Maximum area: finding dimensions that make an area as large as possible.
• Revenue optimization: finding the price or number of items that gives the greatest revenue.
• Projectile kinematics: finding the maximum height of an object and when it occurs.

5. Writing a quadratic model from a word problem

Most optimization problems are not given directly as \(ax^2+bx+c\). You usually need to build the equation first.

To do that, look for:

• what variable to use,
• what quantity must be maximized or minimized,
• any relationships or constraints in the problem.

For example, if a rectangle has perimeter 40 m, and width is \(x\), then:

$$2l+2w=40$$

$$2l+2x=40$$

$$l=20-x$$

If area is the quantity to maximize, then

$$A=x(20-x)=20x-x^2$$

Now the problem has become a quadratic optimization problem.

Worked Example 1: Maximum area of a rectangle

Problem: A rectangular garden has 24 m of fencing. What dimensions give the maximum area?

Step 1: Define a variable.

Let the width be \(x\) m.

Step 2: Use the perimeter.

Since

$$2l+2w=24$$

we get

$$2l+2x=24$$

$$l=12-x$$

Step 3: Write the area function.

$$A=x(12-x)$$

$$A=12x-x^2$$

Step 4: Identify the vertex.

Here, \(a=-1\) and \(b=12\).

$$x=-\frac{b}{2a}=-\frac{12}{2(-1)}=6$$

Step 5: Find the length.

$$l=12-6=6$$

Step 6: Find the maximum area.

$$A=6\cdot 6=36$$

Answer: The maximum area is 36 m², and it occurs when the garden is 6 m by 6 m.

Important idea: For a fixed perimeter, the rectangle with maximum area is a square.

Worked Example 2: Open-top box optimization

Problem: A square sheet of cardboard has side length 20 cm. Squares of side length \(x\) cm are cut from each corner, and the sides are folded up to make an open box. What value of \(x\) gives the maximum volume?

Step 1: Describe the dimensions.

After cutting and folding:

• height = \(x\)
• length = \(20-2x\)
• width = \(20-2x\)

Step 2: Write the volume.

$$V=x(20-2x)^2$$

Step 3: Expand the expression.

First,

$$ (20-2x)^2=400-80x+4x^2 $$

So

$$V=x(400-80x+4x^2)=400x-80x^2+4x^3$$

This expression is cubic, not quadratic, so it is not a quadratic optimization model.

Why include this example? Because it is important to recognize that not every optimization problem is quadratic. In this lesson, we focus on situations that simplify to a quadratic.

Quadratic version of the box idea: Suppose the height is fixed at 2 cm and only the base side length changes. Then volume might take the form

$$V=2x(20-x)=40x-2x^2$$

which is quadratic and can be optimized using the vertex.

Now optimize:

Here, \(a=-2\), \(b=40\).

$$x=-\frac{40}{2(-2)}=10$$

Then

$$V=40(10)-2(10)^2=400-200=200$$

Answer: The maximum volume is 200 cm³ when \(x=10\) cm.

Worked Example 3: Revenue optimization

Problem: A school sells tickets for a play. If the ticket price is \(\$10\), then 300 tickets are sold. For every \(\$1\) increase in price, 20 fewer tickets are sold. What ticket price gives the maximum revenue?

Step 1: Define the variable.

Let \(x\) be the number of \(\$1\) increases.

Then:

• new price = \(10+x\)
• number of tickets sold = \(300-20x\)

Step 2: Write the revenue function.

Revenue = price × quantity

$$R=(10+x)(300-20x)$$

Step 3: Expand.

$$R=3000-200x+300x-20x^2$$

$$R=-20x^2+100x+3000$$

Step 4: Find the vertex.

Here, \(a=-20\), \(b=100\).

$$x=-\frac{100}{2(-20)}=2.5$$

Step 5: Interpret the result.

This means the price should increase by \(\$2.50\).

So the best ticket price is

$$10+2.5=12.5$$

Step 6: Find the maximum revenue.

$$R=-20(2.5)^2+100(2.5)+3000$$

$$R=-20(6.25)+250+3000$$

$$R=-125+250+3000=3125$$

Answer: The maximum revenue is \(\$3125\), achieved when the ticket price is \(\$12.50\).

Worked Example 4: Projectile kinematics

Problem: A ball is thrown upward from a platform. Its height in metres after \(t\) seconds is given by

$$h(t)=-5t^2+20t+15$$

Find the maximum height and when it occurs.

Step 1: Recognize the quadratic model.

The function is already in quadratic form:

$$h(t)=-5t^2+20t+15$$

Since \(a=-5<0\), the parabola opens downward, so the vertex gives the maximum height.

Step 2: Find the time at the vertex.

$$t=-\frac{b}{2a}=-\frac{20}{2(-5)}=2$$

Step 3: Find the height at that time.

$$h(2)=-5(2)^2+20(2)+15$$

$$h(2)=-20+40+15=35$$

Answer: The ball reaches a maximum height of 35 m after 2 seconds.

6. Using vertex form

Sometimes a quadratic is written in vertex form:

$$f(x)=a(x-h)^2+k$$

In this form, the vertex is directly visible as \((h,k)\).

• If \(a>0\), the minimum value is \(k\).
• If \(a<0\), the maximum value is \(k\).

For example, if

$$R(x)=-2(x-4)^2+50$$

then the vertex is \((4,50)\), so the maximum revenue is 50 when \(x=4\).

7. Completing the square as another method

If a quadratic is in standard form, you can sometimes rewrite it in vertex form by completing the square.

For example,

$$f(x)=-x^2+6x+2$$

Factor out the negative from the quadratic terms:

$$f(x)=-(x^2-6x)+2$$

Complete the square:

$$f(x)=-(x^2-6x+9)+2+9$$

$$f(x)=-(x-3)^2+11$$

So the vertex is \((3,11)\), and the maximum value is 11.

8. How to check if your answer makes sense

After solving, always check the result in context.

• Are the dimensions positive?
• Is the time realistic?
• Can the number of items sold be negative? No.
• Does the value of the variable fit the problem conditions?

For example, in a rectangle problem, if you got \(x=15\) but the total perimeter only allows side lengths smaller than 12, then your answer would not make sense.

9. Common mistakes to avoid

• Using the wrong variable: Define clearly what \(x\) represents.
• Forgetting the constraint: Use the given condition, such as perimeter or changing demand.
• Optimizing the wrong expression: Make sure you maximize or minimize the correct quantity, such as area, revenue, or height.
• Not interpreting the vertex correctly: The x-coordinate gives the input value; the y-coordinate gives the maximum or minimum output.
• Ignoring realistic values: In applied problems, not every algebraic answer is meaningful.

10. Problem-solving guide

When you see an optimization word problem, follow this plan:

1. Read carefully and identify what needs to be maximized or minimized.
2. Choose a variable.
3. Use the given information to express all other quantities in terms of that variable.
4. Write the function to optimize.
5. Simplify into quadratic form if possible.
6. Use the vertex formula or vertex form.
7. State the answer with units and context.

11. Quick practice questions

Try these on your own:

1. A rectangle has perimeter 30 m. What dimensions give the maximum area?
2. A product sells for \(\$15\), and 200 units are sold. For each \(\$1\) increase in price, 10 fewer units are sold. What price gives maximum revenue?
3. A ball’s height is given by \(h(t)=-4.9t^2+19.6t+2\). What is its maximum height?

12. Final summary

Quadratic optimization models help you solve real-world problems where a quantity reaches a highest or lowest value. The most important feature of a quadratic is its vertex, since the vertex gives the maximum or minimum value.

To solve these problems, define a variable, build a quadratic equation for the quantity being optimized, and use

$$x=-\frac{b}{2a}$$

to find the turning point. Then interpret your result carefully in the context of the problem.