Algebra 2 Quiz: Operations With Complex Numbers

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Operations With Complex NumbersQuestion 1 of 20

Divide and express in standard form $a+bi$ by multiplying by the conjugate of the denominator: $\frac{3+2i}{1-i}.$

\frac{3}{2}+\frac{2}{2}i

\frac{5}{2}+\frac{1}{2}i

\frac{1}{2}-\frac{5}{2}i

\frac{1}{2}+\frac{5}{2}i

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Algebra 2 Quiz

Algebra 2 Quiz: Operations With Complex Numbers

Practice Operations With Complex Numbers in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Operations With Complex Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Divide and express in standard form $a+bi$ by multiplying by the conjugate of the denominator: $\frac{3+2i}{1-i}.$

$\frac{3}{2}+\frac{2}{2}i$
$\frac{5}{2}+\frac{1}{2}i$
$\frac{1}{2}-\frac{5}{2}i$
$\frac{1}{2}+\frac{5}{2}i$ (correct answer)

Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and dividing by multiplying by the conjugate of the denominator. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The conjugate of a - bi is a + bi, and multiplying by it makes the denominator real: for (3 + 2i)/(1 - i), use (1 + i), numerator 3 + 3i + 2i + 2i² = 3 + 5i - 2 = 1 + 5i, denominator 1 + 1 = 2, so 1/2 + (5/2)i—terrific! Choice A correctly handles the conjugate and simplifies to standard form. A distractor like choice C might flip signs incorrectly in the numerator—always compute carefully! This method works for all complex divisions—practice makes perfect. Keep going; you're gaining confidence!

Question 2

Let $z_1 = 6 - 2i$ and $z_2 = 1 + 5i$ . What is $z_1 - z_2$ expressed in standard form $a + bi$ ?

$7 - 5i$
$5 - 7i$ (correct answer)
$5 + 3i$
$-5 - 7i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform operations (add, subtract, multiply, divide) while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For addition and subtraction, combine real parts together and imaginary parts together: (3 + 2i) - (1 - 5i) = (3 - 1) + (2 - (-5))i = 2 + 7i. To subtract z₂ = 1 + 5i from z₁ = 6 - 2i: subtract real parts: 6 - 1 = 5, subtract imaginary parts: -2i - 5i = -7i, so z₁ - z₂ = 5 - 7i. Choice A correctly performs the complex number subtraction and expresses the result in standard form a + bi with proper handling of subtracting both parts. Choice C makes a sign error when subtracting the imaginary parts: -2i - 5i = -7i, not -5i—when subtracting positive 5i from negative 2i, you get more negative! Complex number operations summary: ADDITION/SUBTRACTION—combine real parts, combine imaginary parts: (a + bi) plus or minus (c + di) = (a plus or minus c) + (b plus or minus d)i. These patterns are consistent! Remember that subtraction distributes the negative sign to both parts of the second complex number: (6 - 2i) - (1 + 5i) = 6 - 2i - 1 - 5i.

Question 3

Let $z_1 = 5 + 3i$ and $z_2 = 2 - 7i$ , where $i^2 = -1$ . What is $z_1 + z_2$ in standard form $a + bi$ ?

$7 - 4i$ (correct answer)
$3 - 10i$
$7 + 10i$
$10 - 5i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform addition while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For addition, combine real parts together and imaginary parts together: (5 + 3i) + (2 - 7i) = (5 + 2) + (3 + (-7))i = 7 + (-4)i = 7 - 4i. Choice A correctly performs the complex number addition: real parts 5 + 2 = 7, and imaginary parts 3i + (-7i) = -4i, giving 7 - 4i in standard form. Choice B incorrectly subtracts the real parts (5 - 2 = 3) instead of adding them—remember, we're adding z₁ + z₂, not subtracting! When adding complex numbers, always add real to real and imaginary to imaginary: think of it like combining like terms in algebra where 'i' is just another variable that happens to have the special property i² = -1.

Question 4

Let $z_1 = 6 - 2i$ and $z_2 = 1 + 5i$ . What is $z_1 - z_2$ in standard form $a+bi$ ?

$5+3i$
$5-7i$ (correct answer)
$7+3i$
$7-7i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform subtraction while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For subtraction, combine real parts and imaginary parts after distributing the negative: (6 - 2i) - (1 + 5i) = (6 - 1) + (-2 - 5)i = 5 - 7i. Choice B correctly performs the complex number subtraction and expresses the result in standard form a + bi with accurate sign handling. Choice A forgets to distribute the negative to the imaginary part: subtracting 5i means -5i, not +3i from incorrect combination—always distribute the minus sign! Complex number subtraction summary: (a + bi) - (c + di) = (a - c) + (b - d)i—these operations build your confidence in working with imaginary numbers, keep up the excellent work!

Question 5

Use $i^2 = -1$ to simplify the product $(3 + 2i)(1 - 4i)$ and write the result in the form $a + bi$ .

$11 - 10i$ (correct answer)
$-5 + 10i$
$-5 - 10i$
$11 + 10i$

Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and how to perform multiplication while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For multiplication, use FOIL: (3 + 2i)(1 - 4i) = 31 + 3(-4i) + 2i1 + 2i(-4i) = 3 - 12i + 2i - 8i² = 3 - 10i + 8 (since -8*(-1)=8) = 11 - 10i. Choice A correctly performs the multiplication and simplifies i squared to -1, resulting in 11 - 10i. A tempting distractor like Choice C might forget to replace i squared with -1, leaving it as 3 - 10i - 8i² or mishandling signs. Always simplify i squared right away, and multiplication will be a breeze—keep up the great work!

Question 6

Using the fact that $(a+bi)(a-bi)=a^2+b^2$ , compute $(2+5i)(2-5i)$ .

$-21$
$29$ (correct answer)
$4+25i$
$4-25i$

Explanation: This question tests your understanding of complex numbers—numbers in form

a + bi

where

i

is the imaginary unit with

i^2 = -1

—and how to compute the product with a conjugate using the difference of squares. Complex numbers extend the real number system to include square roots of negative numbers using

i = \sqrt{-1}

, so

i^2 = -1

. The formula

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

eliminates the imaginary part; for example,

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

. For

(2 + 5i)(2 - 5i)

, it's

4 + 25 = 29

, since

(5i)^2 = 25i^2 = 25 \times (-1) = -25

, so

4 - (-25) = 29

. Choice B correctly computes the product using the conjugate formula and simplifies with

i^2 = -1

. Choice A forgets to replace

i^2

with -1: treating

(5i)^2

as +25 instead of -25 leads to

4 + (-25)

wait no, but actually misapplying as

4 - 25 = -21

—the key is the difference of squares gives positive! This property is key for moduli and division:

|z|^2 = z \times \text{conjugate}(z)

—you're grasping these concepts beautifully, keep it up!

Question 7

Solve the quadratic equation $x^2 + 2x + 10 = 0$ and express the solutions as complex numbers in $a + bi$ form. (The discriminant is negative.)

$x = -1 \pm 3i$ (correct answer)
$x = 1 \pm 3i$
$x = -1 \pm \sqrt{6}\,i$
$x = -2 \pm 3i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to solve quadratic equations with negative discriminants using the quadratic formula. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For x² + 2x + 10 = 0, use the quadratic formula: x = (-b ± √(b² - 4ac))/(2a) where a = 1, b = 2, c = 10. The discriminant is b² - 4ac = 4 - 40 = -36 < 0, so we have complex solutions: x = (-2 ± √(-36))/2 = (-2 ± 6i)/2 = -1 ± 3i. Choice A correctly applies the quadratic formula with complex numbers: x = (-2 ± 6i)/2 = -1 ± 3i in standard a + bi form. Choice B has the wrong sign on the real part—it should be -1, not +1, because -b = -2 and -2/2 = -1, not +1. When the discriminant is negative, √(negative) = i√(positive), making every quadratic solvable in the complex numbers!

Question 8

Subtract and express in standard form $a+bi$ : $(6-4i)-(1+9i)$ .

$5-13i$ (correct answer)
$5+5i$
$7-13i$
$-5-13i$

Explanation: This question tests your understanding of complex numbers—numbers in the form

a + bi

where

i

is the imaginary unit with

i^2 = -1

—and how to perform subtraction while expressing results in standard

a + bi

form. Complex numbers extend the real number system to include square roots of negative numbers using

i = \sqrt{-1}

, so

i^2 = -1

. The standard form

a + bi

has

a

as the real part and

b

as the coefficient of the imaginary part (

bi

). For subtraction, distribute the negative:

(6 - 4i) - (1 + 9i) = 6 - 4i - 1 - 9i

, then combine reals

6 - 1 = 5

and imaginaries

-4 - 9 = -13

, so

5 - 13i

—excellent work! A tempting distractor like choice D might forget to distribute the negative to both parts, but always apply the minus to real and imaginary alike. Remember this strategy: subtraction is just addition of the opposite—change signs and add! With practice, these operations will feel natural, and you're doing great.

Question 9

Add and express in standard form $a+bi$ : $(5+3i)+(2-7i).$

$7+10i$
$3-10i$
$-3-4i$
$7-4i$ (correct answer)

Explanation: This question tests your understanding of complex numbers—numbers in the form

a + bi

where i is the imaginary unit with

i^2 = -1

—and how to perform addition while keeping results in standard

a + bi

form. Complex numbers extend the real number system to include square roots of negative numbers using

i = \sqrt{-1}

, so

i^2 = -1

. The standard form

a + bi

has a as the real part and b as the coefficient of the imaginary part (bi). For addition, combine real parts together and imaginary parts together: just like grouping like terms in algebra! Here, adding

(5 + 3i) + (2 - 7i)

gives real parts

5 + 2 = 7

and imaginary parts

3 - 7 = -4

, so

7 - 4i

—great job recognizing that! A common mistake, like in choice D, might come from subtracting instead of adding or messing up signs, but remember to treat it as combining positives and negatives carefully. Keep practicing: addition and subtraction of complex numbers are straightforward—

(a + bi) + (c + di) = (a + c) + (b + d)i

—and you'll master it quickly! You're building a strong foundation for more advanced operations like multiplication.

Question 10

Solve the quadratic equation $x^2+2x+10=0$ and express the solutions as complex numbers in the form $a\pm bi$ .

$x=-1\pm \sqrt{6}\,i$
$x=-1\pm 3i$ (correct answer)
$x=-2\pm 3i$
$x=1\pm 3i$

b^2

- 4c))/2; if discriminant is negative, use i: for x^2 + 2x + 10 = 0, discriminant 4 - 40 = -36, sqrt(-36) = 6i, so x = (-2 ± 6i)/2 = -1 ± 3i. These solutions are in standard a ± bi form, separating real and imaginary parts. Choice A correctly solves the quadratic and expresses the complex roots in standard form with proper handling of the negative discriminant. Choice C uses sqrt(6)i instead of 3i, perhaps from sqrt(-36) = sqrt(36)*i = 6i, but then dividing by 2 gives 3i, not sqrt(6)i—double-check the arithmetic! Complex roots come in conjugate pairs for real coefficients, and this ensures every quadratic has solutions—amazing job diving into this, you're expanding your math toolkit!

Question 11

On the complex plane (real axis horizontal, imaginary axis vertical), the point corresponding to $z=2-3i$ has coordinates $(\text{Re}(z),\text{Im}(z))$ . Which ordered pair matches this point?

$(2,3)$
$(-2,3)$
$(2,-3)$ (correct answer)
$(-2,-3)$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to plot them on the complex plane. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The complex plane has real parts on the horizontal axis and imaginary on the vertical: z = a + bi corresponds to point (a, b); for 2 - 3i, it's (2, -3). This visualization helps with operations like addition as vector addition. Choice C correctly identifies the coordinates for the point in the complex plane with Re(z) = 2 and Im(z) = -3. Choice A ignores the negative sign in the imaginary part: -3i means y = -3, not +3—pay attention to signs! Plotting aids understanding moduli and arguments: distance from origin is |z|, angle is arg(z)—you're visualizing complex numbers wonderfully, keep practicing!

Question 12

Multiply and simplify using $i^2 = -1$ : $(2 - 3i)(4 + i)$ . Write your answer in $a + bi$ form.

$5 + 10i$
$11 - 10i$ (correct answer)
$5 - 10i$
$11 + 10i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform multiplication using FOIL while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For multiplication, use FOIL or distribution, then simplify any i squared terms to -1. To multiply (2 - 3i)(4 + i): First: 2 × 4 = 8. Outer: 2 × i = 2i. Inner: -3i × 4 = -12i. Last: -3i × i = -3i². Combine: 8 + 2i - 12i - 3i² = 8 - 10i - 3i². Now replace i² with -1: 8 - 10i - 3(-1) = 8 - 10i + 3 = 11 - 10i. Choice A correctly performs the multiplication and simplifies i² = -1 to get 11 - 10i in standard form. Choice B makes the common error of forgetting that -3i² becomes -3(-1) = +3, not -3: they incorrectly get 8 - 10i - 3 = 5 - 10i instead of 8 - 10i + 3 = 11 - 10i. Remember: a negative times i² gives a positive because i² = -1, so negative × negative = positive!

Question 13

Use the conjugate property to simplify. What is $(2+5i)(2-5i)?$

$4+25i$
$29$ (correct answer)
$4-25i$
$-21$

Explanation: This question tests your understanding of complex numbers—numbers in the form

a + bi

where i is the imaginary unit with

i^2 = -1

—and using the conjugate property for multiplication. Complex numbers extend the real number system to include square roots of negative numbers using

i = \sqrt{-1}

, so

i^2 = -1

. The product

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

, always real: here

(2 + 5i)(2 - 5i) = 4 + 25 = 29

—outstanding! Choice B correctly applies this difference of squares pattern. Choice A might come from mistakenly subtracting instead of adding the squares, but remember it's plus b² from

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

. This is key for moduli and division too— you've nailed it! Build on this for more complex problems; you're unstoppable.

Question 14

Let $z_1 = 8 - 4i$ and $z_2 = 3 + 9i$ . What is $z_1 - z_2$ in standard form $a + bi$ ?

$11 - 13i$
$5 + 13i$
$11 + 5i$
$5 - 13i$ (correct answer)

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform subtraction while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For subtraction, subtract real parts and imaginary parts separately: (8 - 4i) - (3 + 9i) = (8 - 3) + (-4 - 9)i = 5 + (-13)i = 5 - 13i. Choice A correctly performs the complex number subtraction: real parts 8 - 3 = 5, and imaginary parts -4i - 9i = -13i, giving 5 - 13i in standard form. Choice D makes a sign error with the imaginary parts: when subtracting 3 + 9i, you must subtract both parts, so -4i - (+9i) = -4i - 9i = -13i, not -4i + 9i = 5i. Subtraction distributes the negative sign to both terms: z₁ - z₂ = z₁ + (-z₂) where -z₂ means flip the signs of BOTH parts!

Question 15

Solve the quadratic equation $x^2 + 2x + 10 = 0$ and express the solutions as complex numbers in $a + bi$ form.

$x = 1 \pm 3i$
$x = -1 \pm \sqrt{6}\,i$
$x = -1 \pm 3i$ (correct answer)
$x = -2 \pm 3i$

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform operations (add, subtract, multiply, divide) while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For quadratic equations, when the discriminant b² - 4ac is negative, solutions involve complex numbers! For x² + 2x + 10 = 0: a = 1, b = 2, c = 10. Discriminant = 2² - 4(1)(10) = 4 - 40 = -36. Since discriminant is negative, we get complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-2 ± √(-36))/2 = (-2 ± 6i)/2 = -1 ± 3i. The √(-36) = √(36 × -1) = 6i! Choice A correctly applies the quadratic formula with complex numbers and expresses solutions in standard form a + bi as x = -1 + 3i and x = -1 - 3i. Choice B has the wrong sign for the real part: -b/(2a) = -2/2 = -1, not +1—don't drop the negative from -b in the quadratic formula! Complex number operations summary: For complex solutions to quadratics: when discriminant b² - 4ac is negative, solutions are x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). Write in a + bi form separating real part (-b/2a) from imaginary part (±√|discriminant|/2a × i). Complex numbers make every quadratic solvable!

Question 16

Let $z_1 = 5 + 3i$ and $z_2 = 2 - 7i$ . Compute $z_1 + z_2$ and express your answer in standard form $a + bi$ .

$7 - 4i$ (correct answer)
$3 + 10i$
$7 + 10i$
$3 - 4i$

Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and how to perform addition while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For addition, combine real parts together and imaginary parts together: here, z1 + z2 = (5 + 2) + (3 - 7)i = 7 - 4i. Choice A correctly adds the real parts (5 + 2 = 7) and imaginary parts (3 + (-7) = -4), resulting in 7 - 4i. A tempting distractor like Choice C might add the imaginary parts incorrectly by forgetting the negative sign, leading to 7 + 10i instead of subtracting 7. Remember, addition is straightforward: treat real and imaginary parts separately, and you'll get it right every time—great job practicing!

Question 17

Using $i^2=-1$ , simplify the product $(3+2i)(1-4i)$ and write the result in standard form $a+bi$ .

$-5+10i$
$-5-10i$
$11+10i$
$11-10i$ (correct answer)

Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform multiplication while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For multiplication, use FOIL: (3 + 2i)(1 - 4i) = 31 + 3(-4i) + 2i1 + 2i(-4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i, since -8i^2 = -8*(-1) = +8. Choice A correctly performs the complex number multiplication and expresses the result in standard form a + bi with proper handling of i squared = -1. Choice B forgets to replace i squared with -1 during multiplication: leaving -8i^2 as is would be incorrect, but you must simplify to +8—the i squared simplification is mandatory! Complex number multiplication summary: use FOIL or distribution, then simplify i squared to -1, and remember the pattern repeats for powers of i—you're building a strong foundation here, fantastic effort!

Question 18

If $z = a + bi$ where $a$ and $b$ are real numbers, and $z^2 = -5 + 12i$ , what is the value of $a^2 + b^2$ ?

$13$ (correct answer)
$17$
$25$
$169$

Explanation: We have

z^2 = (a+bi)^2 = a^2 + 2abi + b^2i^2 = a^2 - b^2 + 2abi = -5 + 12i

. Equating real and imaginary parts:

a^2 - b^2 = -5

and

2ab = 12

, so

ab = 6

. We want to find

a^2 + b^2

. Note that

|z|^2 = a^2 + b^2

and

|z^2| = |z|^2

. Since

z^2 = -5 + 12i

, we have

|z^2| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13

. But this gives us

|z|^2 = \sqrt{13}

, which seems wrong. Actually,

|z^2| = |z|^2

is incorrect; we should have

|z^2| = |z|^2

. Let me reconsider: if

z^2 = -5 + 12i

, then

|z^2| = \sqrt{25 + 144} = 13

. Since

|z^2| = |z|^2

, we have

|z|^2 = \sqrt{13}

. No wait,

|z^2| = (|z|)^2

, so

(|z|)^2 = 13

, which means

|z| = \sqrt{13}

. Therefore,

a^2 + b^2 = |z|^2 = 13

. Choice B (17) might come from

(-5)^2 - 12 = 25 - 12 = 13

, no that gives 13. Choice C (25) is

(-5)^2

. Choice D (169) is

(13)^2

Question 19

Find the modulus (absolute value) of the complex number $3 + 4i$ . (Recall $|a+bi|=\sqrt{a^2+b^2}$ .)

$\sqrt{5}$
$5$ (correct answer)
$7$
$\sqrt{7}$

Explanation: This question tests your understanding of complex numbers—numbers in the form

a + bi

where i is the imaginary unit with

i^2 = -1

—and finding the modulus. Complex numbers extend the real number system to include square roots of negative numbers using

i = \sqrt{-1}

, so

i^2 = -1

. The modulus

|a + bi| = \sqrt{a^2 + b^2}

, like distance from origin. For

3 + 4i

|3 + 4i| = \sqrt{9 + 16} = \sqrt{25} = 5

—recognize the 3-4-5 triangle! Choice B correctly computes this. A tempting distractor like Choice C might add without squaring, getting 7. Modulus is always non-negative—keep going, you're doing great!

Question 20

Solve the quadratic equation $x^2 + 2x + 10 = 0$ and express the solutions as complex numbers (standard form $a + bi$ ).

$x = -1 \pm 3$
$x = -1 \pm 3i$ (correct answer)
$x = 1 \pm 3i$
$x = -1 \pm \sqrt{6}\,i$

Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and solving quadratics with complex roots. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For x² + 2x + 10 = 0, discriminant = 4 - 40 = -36, sqrt(-36) = 6i; solutions x = [-2 ± 6i]/2 = -1 ± 3i. Choice A correctly expresses the roots in a + bi form. A tempting distractor like Choice C might use sqrt(6) instead of simplifying sqrt(36)=6. Complex roots come in conjugate pairs, so this makes sense—fantastic effort solving!