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Intermediate Algebra / LEVEL 1 · DIFFICULTY 1/5

Complex Numbers and Quadratics

Extend arithmetic and interpret quadratic roots.

3 stages · 24 practice problems · two 6-question assessment forms

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MINI QUEST IA 1.1Multiplying Complex NumbersRead the lesson
MINI QUEST IA 1.2Modulus and ConjugationRead the lesson
MINI QUEST IA 1.3DiscriminantsRead the lesson
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STAGE IA 1.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Multiplying Complex Numbers

Goal: Understand and apply multiplying complex numbers.

Before you begin: Linear and quadratic equations, factoring and introductory functions.

Understand the idea

Complex numbers obey distributive multiplication with the additional rule $\displaystyle i^{2} = -1$ . Products of two imaginary terms contribute to the real part, so both constant products matter.

(a+bi)(c+di)=

\displaystyle \left(ac - bd\right) + \left(ad + bc\right)\, i

Choose and carry out a method

Expand all four products, replace $\displaystyle i^{2}$ by −1, and group real and imaginary terms. Enter only the component requested.

Check the reasoning

Multiplying by i rotates a point by a quarter-turn. A second multiplication by i negates it, consistent with $\displaystyle i^{2} = -1$ .

WORKED EXAMPLE 1

Find the real part of $\displaystyle \left(3 + 6\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .

Expand the product and replace $\displaystyle i^{2}$ by -1.
Real terms are $\displaystyle 2 \cdot 3 + 3 \cdot 6 \cdot \left(-1\right)$ .
The real part is -12; the cross-terms belong to the imaginary part.

-12

WORKED EXAMPLE 2

Find the real part of $\displaystyle \left(4 + 7\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .

Expand the product and replace $\displaystyle i^{2}$ by -1.
Real terms are $\displaystyle 2 \cdot 4 + 3 \cdot 7 \cdot \left(-1\right)$ .
The real part is -13; the cross-terms belong to the imaginary part.

-13

Common pitfalls

Possible mix-up: The real part is only ac.

The product bi·di contributes −bd to the real part.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why does multiplying two imaginary terms produce a real term?

Preview the eight practice prompts

Problem 1
Find the real part of $\displaystyle \left(6 + 9\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
Practise this exact problem →
Problem 2
Find the real part of $\displaystyle \left(7 + 10\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
Practise this exact problem →
Problem 3
Find the real part of $\displaystyle \left(8 + 11\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
Practise this exact problem →
Problem 4
Find the real part of $\displaystyle \left(9 + 12\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
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Problem 5
Find the real part of $\displaystyle \left(10 + 13\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
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Problem 6
Find the real part of $\displaystyle \left(11 + 14\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .
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Problem 7New context
Two planar transformations are encoded by complex factors $\displaystyle 12 + 15\, i$ and $\displaystyle 2 + 3\, i$ . Their composition uses the product. What is its real part?
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Problem 8New context
Two planar transformations are encoded by complex factors $\displaystyle 13 + 16\, i$ and $\displaystyle 2 + 3\, i$ . Their composition uses the product. What is its real part?
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Open stage IA 1.1 in the student workspace →

STAGE IA 1.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Modulus and Conjugation

Useful preparation: Multiplying Complex Numbers

Goal: Understand and apply modulus and conjugation.

Before you begin: Multiplying Complex Numbers

Understand the idea

The modulus of x+yi is its distance from the origin in the complex plane. Multiplying a number by its conjugate eliminates the imaginary cross terms and yields the squared distance.

|x+yi|²=(x+yi)(x−yi)=

\displaystyle x^{2} + y^{2}

Choose and carry out a method

Square both real coordinates and add. If the question asks for the squared modulus, stop before taking a square root.

Check the reasoning

The result is nonnegative, even for negative coordinates. Conjugation preserves distance while reflecting the point across the real axis.

WORKED EXAMPLE 1

Find $\displaystyle \left|3 - 5\, i\right|^{2}$ .

Multiplying a complex number by its conjugate gives its squared modulus.
$\displaystyle \left(3 - 5\, i\right)\, \left(3 + 5\, i\right) = 3^{2} + 5^{2}$ .
The squared distance is 34, which is real and nonnegative.

WORKED EXAMPLE 2

Find $\displaystyle \left|4 - 6\, i\right|^{2}$ .

Multiplying a complex number by its conjugate gives its squared modulus.
$\displaystyle \left(4 - 6\, i\right)\, \left(4 + 6\, i\right) = 4^{2} + 6^{2}$ .
The squared distance is 52, which is real and nonnegative.

Common pitfalls

Possible mix-up: Square the imaginary unit as part of the distance.

The coordinate y is squared; squared distance is $\displaystyle x^{2} + y^{2}$ .

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain geometrically why conjugate numbers have equal modulus.

Preview the eight practice prompts

Problem 1
Find $\displaystyle \left|6 - 8\, i\right|^{2}$ .
Practise this exact problem →
Problem 2
Find $\displaystyle \left|7 - 9\, i\right|^{2}$ .
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Problem 3
Find $\displaystyle \left|8 - 10\, i\right|^{2}$ .
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Problem 4
Find $\displaystyle \left|9 - 11\, i\right|^{2}$ .
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Problem 5
Find $\displaystyle \left|10 - 12\, i\right|^{2}$ .
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Problem 6
Find $\displaystyle \left|11 - 13\, i\right|^{2}$ .
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Problem 7New context
A displacement is represented by the complex number $\displaystyle 12 - 14\, i$ . What is its squared distance from the origin?
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Problem 8New context
A displacement is represented by the complex number $\displaystyle 13 - 15\, i$ . What is its squared distance from the origin?
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Open stage IA 1.2 in the student workspace →

STAGE IA 1.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Discriminants

Useful preparation: Modulus and Conjugation

Goal: Understand and apply discriminants.

Before you begin: Modulus and Conjugation

PARENT EQUATION

Quadratic parent: $\displaystyle y = x^{2}$

A U-shaped curve with vertex (0, 0), symmetric about the y-axis.

How does the question’s equation change this graph?

Write $\displaystyle y = a\, \left(x - h\right)^{2} + k$ . Move the vertex to (h, k), stretch vertically by $\displaystyle \left|a\right|$ , and reflect across the x-axis if $\displaystyle a < 0$ . In $\displaystyle ax^{2} + bx + c$ , $\displaystyle h = \frac{-b}{2\, a}$ and $\displaystyle k = c - \frac{b^{2}}{4\, a}$ .

Check: The sign inside the brackets is opposite the horizontal shift: $\displaystyle \left(x - 3\right)^{2}$ moves right 3. This form requires $\displaystyle a \ne 0$ .

First identify this family. Then read the coefficients and shifts in your actual equation; this reference does not replace the question’s given values.

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: $\displaystyle y = x^{2}$

Start with $\displaystyle y = x^{2}$ . Complete the square: $\displaystyle y = 1\, \left(x - \left(1\right)\right)^{2} + \left(-4\right)$ . The new vertex is (1, -4). The first coordinate gives the horizontal shift and the second gives the vertical shift. Multiply vertical distances from the vertex by 1.

Transform the parent graph

Teal moving graph: $\displaystyle y = 1\, x^{2} + \left(0\right)\, x + \left(0\right)$

Blue: $\displaystyle y = 1\, x^{2} - 2\, x - 3$

Moving point: (-4, 16)

Teal shows the current transformation. Blue is the target. At $\displaystyle 100\%$ they coincide; intermediate graphs are learning frames, not additional solutions.

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Curve coefficient a: 1

Coefficient b: -2

Vertical intercept c: -3

Choose the point’s x: 1

Point (1, -4)

\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right) = -4

Why does it bend?

The $\displaystyle x^{2}$ term makes the change in y vary with x. This parabola opens upward because a is positive.

Three points made by the blue rule
x	Substitute into the rule	y
-1	$\displaystyle 1 \times \left(-1\right)^{2} + \left(-2\right) \times \left(-1\right) + \left(-3\right)$	0
0	$\displaystyle 1 \times \left(0\right)^{2} + \left(-2\right) \times \left(0\right) + \left(-3\right)$	-3
1	$\displaystyle 1 \times \left(1\right)^{2} + \left(-2\right) \times \left(1\right) + \left(-3\right)$	-4

Understand the idea

Completing the square in a quadratic reveals the expression $\displaystyle b^{2} - 4\, ac$ . Its sign controls whether the two roots are distinct real numbers, one repeated real number or a conjugate nonreal pair.

\displaystyle \Delta = b^{2} - 4\, ac

Choose and carry out a method

Identify a, b and c with their signs before substituting. Compute the discriminant using parentheses around a negative coefficient.

Check the reasoning

A perfect-square discriminant may give rational roots when coefficients are integers. A positive value alone guarantees real roots, not integer roots.

WORKED EXAMPLE 1

Find the discriminant of $\displaystyle x^{2} - 5\, x + 3 = 0$ .

For $\displaystyle ax^{2} + bx + c = 0$ , the discriminant is $\displaystyle b^{2} - 4\, ac$ .
$\displaystyle D = \left(-5\right)^{2} - 4 \cdot 1 \cdot 3$ .
$\displaystyle D = 13$ , so this example has two distinct real roots.

WORKED EXAMPLE 2

Find the discriminant of $\displaystyle x^{2} - 7\, x + 4 = 0$ .

For $\displaystyle ax^{2} + bx + c = 0$ , the discriminant is $\displaystyle b^{2} - 4\, ac$ .
$\displaystyle D = \left(-7\right)^{2} - 4 \cdot 1 \cdot 4$ .
$\displaystyle D = 33$ , so this example has two distinct real roots.

Common pitfalls

Possible mix-up: A negative b gives a negative $\displaystyle b^{2}$ .

Squaring a signed coefficient gives a nonnegative value.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

What can the discriminant tell you without calculating either root?

Preview the eight practice prompts

Problem 1
Find the discriminant of $\displaystyle x^{2} - 11\, x + 6 = 0$ .
Practise this exact problem →
Problem 2
Find the discriminant of $\displaystyle x^{2} - 13\, x + 7 = 0$ .
Practise this exact problem →
Problem 3
Find the discriminant of $\displaystyle x^{2} - 15\, x + 8 = 0$ .
Practise this exact problem →
Problem 4
Find the discriminant of $\displaystyle x^{2} - 17\, x + 9 = 0$ .
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Problem 5
Find the discriminant of $\displaystyle x^{2} - 19\, x + 10 = 0$ .
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Problem 6
Find the discriminant of $\displaystyle x^{2} - 21\, x + 11 = 0$ .
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Problem 7New context
A projectile-height model reaches the ground when $\displaystyle x^{2} - 23\, x + 12 = 0$ . What is the discriminant used to check whether real crossing times exist?
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Problem 8New context
A projectile-height model reaches the ground when $\displaystyle x^{2} - 25\, x + 13 = 0$ . What is the discriminant used to check whether real crossing times exist?
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Open stage IA 1.3 in the student workspace →

Browse another level →

Complex Numbers and Quadratics

Multiplying Complex Numbers

Understand the idea

Choose and carry out a method

Check the reasoning

Find the real part of (3+6 i) (2+3 i)\displaystyle \left(3 + 6\, i\right)\, \left(2 + 3\, i\right)(3+6i)(2+3i), where i2=−1\displaystyle i^{2} = -1i2=−1.

Find the real part of (4+7 i) (2+3 i)\displaystyle \left(4 + 7\, i\right)\, \left(2 + 3\, i\right)(4+7i)(2+3i), where i2=−1\displaystyle i^{2} = -1i2=−1.

Common pitfalls

Explain it to yourself

Modulus and Conjugation

Understand the idea

Choose and carry out a method

Check the reasoning

Find ∣3−5 i∣2\displaystyle \left|3 - 5\, i\right|^{2}∣3−5i∣2.

Find ∣4−6 i∣2\displaystyle \left|4 - 6\, i\right|^{2}∣4−6i∣2.

Common pitfalls

Explain it to yourself

Discriminants

Change the rule. Watch the graph.

Parent equation: y=x2\displaystyle y = x^{2}y=x2

Understand the idea

Choose and carry out a method

Check the reasoning

Find the discriminant of x2−5 x+3=0\displaystyle x^{2} - 5\, x + 3 = 0x2−5x+3=0.

Find the discriminant of x2−7 x+4=0\displaystyle x^{2} - 7\, x + 4 = 0x2−7x+4=0.

Common pitfalls

Explain it to yourself

Find the real part of $\displaystyle \left(3 + 6\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .

Find the real part of $\displaystyle \left(4 + 7\, i\right)\, \left(2 + 3\, i\right)$ , where $\displaystyle i^{2} = -1$ .

Find $\displaystyle \left|3 - 5\, i\right|^{2}$ .

Find $\displaystyle \left|4 - 6\, i\right|^{2}$ .

Parent equation: $\displaystyle y = x^{2}$

Find the discriminant of $\displaystyle x^{2} - 5\, x + 3 = 0$ .

Find the discriminant of $\displaystyle x^{2} - 7\, x + 4 = 0$ .