Math 2 Quiz: Quadratics With Complex Solutions

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Quadratics With Complex SolutionsQuestion 1 of 13

If the quadratic $ax^2 + bx + c = 0$ has solutions $x = \frac{1 \pm i\sqrt{3}}{2}$ , and $a + b + c = 2$ , what is the value of $a$ ?

a = 1

from sum of coefficients condition

a = 2

from sum of coefficients condition

a = 3

from sum of coefficients condition

a = 4

from sum of coefficients condition

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Math 2 Quiz

Math 2 Quiz: Quadratics With Complex Solutions

Practice Quadratics With Complex Solutions in Math 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 Quadratics With Complex Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Question 1

If the quadratic $ax^2 + bx + c = 0$ has solutions $x = \frac{1 \pm i\sqrt{3}}{2}$ , and $a + b + c = 2$ , what is the value of $a$ ?

$a = 1$ from sum of coefficients condition
$a = 2$ from sum of coefficients condition (correct answer)
$a = 3$ from sum of coefficients condition
$a = 4$ from sum of coefficients condition

Explanation: From the solutions, the sum is

\frac{1 + i\sqrt{3}}{2} + \frac{1 - i\sqrt{3}}{2} = 1

and the product is

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

. So

-\frac{b}{a} = 1

and

\frac{c}{a} = 1

, giving

b = -a

and

c = a

. From

a + b + c = 2

a + (-a) + a = a = 2

. Choices A, C, and D represent common errors in coefficient relationships or arithmetic.

Question 2

A quadratic equation $ax^2 + bx + c = 0$ has solutions $x = 2 \pm 3i$ . If $a = 1$ , what is the value of $b + c$ ?

$9$ (correct answer)
$17$
$-17$
$-9$

Explanation: If the solutions are

x = 2 \pm 3i

, then the quadratic can be written as

(x - (2 + 3i))(x - (2 - 3i)) = 0

. Expanding:

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

. So

b = -4

and

c = 13

, giving

b + c = 9

Question 3

The complex number $z = 2 - 3i$ is a root of the quadratic equation $x^2 + px + q = 0$ where $p$ and $q$ are real. What is another quadratic equation with real coefficients that has $z^2$ as a root?

$x^2 - 5x + 13 = 0$
$x^2 - 10x + 169 = 0$
$x^2 + 5x + 13 = 0$
$x^2 + 10x + 169 = 0$ (correct answer)

Explanation: When you see a quadratic with complex roots, remember that complex roots of polynomials with real coefficients always come in conjugate pairs. Since

z = 2 - 3i

is a root, its conjugate

\overline{z} = 2 + 3i

must also be a root. To find a quadratic with

z^2

as a root, first calculate

z^2

z^2 = (2 - 3i)^2 = 4 - 12i + 9i^2 = 4 - 12i - 9 = -5 - 12i

Since we need real coefficients, the conjugate

\overline{z^2} = -5 + 12i

must also be a root. For a quadratic with roots

r_1

and

r_2

, the equation is

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

. Sum of roots:

(-5 - 12i) + (-5 + 12i) = -10

Product of roots:

(-5 - 12i)(-5 + 12i) = 25 - (12i)^2 = 25 + 144 = 169

Therefore:

x^2 - (-10)x + 169 = x^2 + 10x + 169 = 0

Answer choice A has the wrong sum of roots (5 instead of -10) and wrong product (13 instead of 169). Choice B has the correct product but wrong sum (-10 instead of +10). Choice C has both the wrong sum and wrong product. Only choice D gives us

x^2 + 10x + 169 = 0

. Study tip: When working with powers of complex numbers that need real coefficient polynomials, always find the conjugate of the powered result and use sum/product formulas to build your quadratic.

Question 4

For what value of $k$ does the equation $x^2 - 6x + k = 0$ have solutions with imaginary part equal to $\pm 4$ ?

$k = 9$
$k = 25$ (correct answer)
$k = 16$
$k = 41$

Explanation: Using the quadratic formula:

x = \frac{6 \pm \sqrt{36 - 4k}}{2} = 3 \pm \frac{\sqrt{36 - 4k}}{2}

. For the solutions to have imaginary part

\pm 4

, we need

36 - 4k < 0

(so the discriminant is negative) and

\frac{\sqrt{4k - 36}}{2} = 4

. This gives us

\sqrt{4k - 36} = 8

, so

4k - 36 = 64

, which means

4k = 100

and

k = 25

. Let's verify: when

k = 25

, we get

x = 3 \pm \frac{\sqrt{-64}}{2} = 3 \pm \frac{8i}{2} = 3 \pm 4i

. Choice A gives imaginary part 0. Choice C gives imaginary part

\pm 2

. Choice D gives imaginary part

\pm 5

Question 5

If $r$ and $s$ are the complex solutions to $x^2 - 4x + 29 = 0$ , what is the value of $r^2 + s^2$ ?

$42$
$-58$
$-42$ (correct answer)
$58$

Explanation: When you encounter complex solutions to a quadratic equation, you can often find relationships between the solutions without actually solving for them explicitly. This problem tests your understanding of Vieta's formulas and algebraic manipulation. For the quadratic

x^2 - 4x + 29 = 0

, you can use Vieta's formulas: the sum of roots

r + s = 4

and the product

rs = 29

. To find

r^2 + s^2

, use the algebraic identity

(r + s)^2 = r^2 + 2rs + s^2

, which rearranges to

r^2 + s^2 = (r + s)^2 - 2rs

. Substituting the values:

r^2 + s^2 = (4)^2 - 2(29) = 16 - 58 = -42

. This confirms answer choice C is correct. Let's examine why the other answers are wrong. Choice A (42) likely comes from incorrectly adding

2rs

instead of subtracting it:

16 + 26 = 42

. Choice B (-58) might result from forgetting to include the

(r + s)^2

term entirely and just using

-2rs = -58

. Choice D (58) could come from computing

2rs = 58

and forgetting about the other terms completely. The key insight here is recognizing that even though the roots are complex, you don't need to find them explicitly. Vieta's formulas combined with algebraic identities let you work with sums and products of roots directly. This approach saves time and reduces calculation errors on exam problems involving complex solutions.

Question 6

The quadratic equation $ax^2 + bx + c = 0$ has solutions $x = 3 \pm 2i$ . If $a = 1$ , what is the value of $b^2 - 4ac$ ?

$-16$ (correct answer)
$16$
$-64$
$64$

Explanation: Since the solutions are

3 + 2i

and

3 - 2i

, we can write the quadratic as

a(x - (3 + 2i))(x - (3 - 2i)) = 0

. Expanding:

a((x - 3) - 2i)((x - 3) + 2i) = a((x - 3)^2 - (2i)^2) = a((x - 3)^2 + 4) = a(x^2 - 6x + 9 + 4) = a(x^2 - 6x + 13)

. With

a = 1

, we have

x^2 - 6x + 13 = 0

, so

b = -6

and

c = 13

. Therefore

b^2 - 4ac = (-6)^2 - 4(1)(13) = 36 - 52 = -16

. Choice B incorrectly takes the absolute value. Choice C squares the imaginary part incorrectly. Choice D combines both errors.

Question 7

The quadratic $3x^2 + 12x + 21 = 0$ has complex solutions. If these solutions are written in the form $a \pm bi$ where $a$ and $b$ are real, what is $a^2 + b^2$ ?

$4$
$7$ (correct answer)
$13$
$16$

Explanation: First divide by 3:

x^2 + 4x + 7 = 0

. Using the quadratic formula:

x = \frac{-4 \pm \sqrt{16 - 28}}{2} = \frac{-4 \pm \sqrt{-12}}{2} = \frac{-4 \pm 2i\sqrt{3}}{2} = -2 \pm i\sqrt{3}

. So

a = -2

and

b = \sqrt{3}

. Therefore

a^2 + b^2 = (-2)^2 + (\sqrt{3})^2 = 4 + 3 = 7

. Choice A uses only

a^2

. Choice C incorrectly calculates

4 + 9 = 13

. Choice D uses

(-4)^2

instead of

(-2)^2

Question 8

If $z$ is a solution to $2x^2 - 8x + 17 = 0$ , what is the value of $|z|^2$ ?

$\frac{17}{2}$ (correct answer)
$\frac{17}{4}$
$\frac{25}{4}$
$\frac{25}{2}$

Explanation: Using the quadratic formula:

z = \frac{8 \pm \sqrt{64 - 136}}{4} = \frac{8 \pm \sqrt{-72}}{4} = \frac{8 \pm 6i\sqrt{2}}{4} = 2 \pm \frac{3i\sqrt{2}}{2}

. For a complex number

a + bi

, we have

|a + bi|^2 = a^2 + b^2

. Here,

a = 2

and

b = \pm\frac{3\sqrt{2}}{2}

, so

|z|^2 = 2^2 + \left(\frac{3\sqrt{2}}{2}\right)^2 = 4 + \frac{9 \cdot 2}{4} = 4 + \frac{18}{4} = 4 + \frac{9}{2} = \frac{8 + 9}{2} = \frac{17}{2}

. Choice B divides by an extra 2. Choice C uses

5^2

instead of calculating correctly. Choice D doubles the correct answer.

Question 9

The equation $(x - h)^2 + k^2 = 0$ has solutions $x = 5 \pm 3i$ . What is the value of $h + k$ ?

$8$ only
$2$ or $-8$
$8$ or $-2$
$2$ or $8$ (correct answer)

Explanation: When you encounter an equation involving complex numbers, you need to work backwards from the given solutions to find the parameters. The equation

(x - h)^2 + k^2 = 0

can be rearranged to

(x - h)^2 = -k^2

, which means

x - h = \pm ki

, so

x = h \pm ki

. Comparing this form with the given solutions

x = 5 \pm 3i

, you can see that

h = 5

and

ki = 3i

, which means

k = 3

k = -3

(since

(-3)i = -3i

and

\pm(-3i) = \mp 3i = \pm 3i

). Therefore,

h + k = 5 + 3 = 8

h + k = 5 + (-3) = 2

. Looking at the answer choices: Choice A gives only 8, missing the possibility that

k = -3

. Choice B incorrectly suggests

h + k

could be 2 or -8, but -8 is impossible since

h = 5

(positive) and even if

k = -3

, the sum is still 2. Choice C suggests 8 or -2, but -2 is impossible for the same reason. Choice D correctly identifies both possible values: 2 and 8. The key insight is recognizing that

k

can be either positive or negative since

k^2

appears in the original equation, making both values equally valid. Always consider that when you see

k^2

in an equation,

k

itself could be positive or negative, leading to multiple valid parameter combinations.

Question 10

If $w = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$ is written in the form $a + bi$ , what quadratic equation with integer coefficients has $w$ and $\bar{w}$ as roots?

$x^2 + x - 1 = 0$
$x^2 + x + 1 = 0$
$x^2 - x - 1 = 0$
$x^2 - x + 1 = 0$ (correct answer)

Explanation: When you see a complex number in polar form and need to find a quadratic equation with that number and its conjugate as roots, you're working with the relationship between complex conjugate pairs and polynomial coefficients. First, let's convert

w = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})

to rectangular form. Since

\cos(\frac{\pi}{3}) = \frac{1}{2}

and

\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}

, we have

w = \frac{1}{2} + i\frac{\sqrt{3}}{2}

. The conjugate is

\bar{w} = \frac{1}{2} - i\frac{\sqrt{3}}{2}

. For a quadratic with roots

w

and

\bar{w}

, we can use the fact that if

r_1

and

r_2

are roots, then the quadratic is

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

. The sum is:

w + \bar{w} = \frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2} = 1

The product is:

w\bar{w} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1

Therefore, our quadratic is

x^2 - 1x + 1 = 0

, or

x^2 - x + 1 = 0

. Choice A gives sum = -1 and product = -1. Choice B gives sum = -1 and product = 1. Choice C gives sum = 1 and product = -1. Only choice D matches our calculated sum of 1 and product of 1. Remember: for complex conjugate pairs, their sum is always real (twice the real part) and their product equals the square of the modulus.

Question 11

If $z_1 = 3 + 4i$ and $z_2 = 3 - 4i$ are the roots of a monic quadratic equation $x^2 + px + q = 0$ , what quadratic equation has roots $2z_1$ and $2z_2$ ?

$x^2 - 12x + 100 = 0$ (correct answer)
$x^2 + 12x + 100 = 0$
$x^2 - 6x + 25 = 0$
$x^2 + 6x + 25 = 0$

Explanation: First,

2z_1 = 6 + 8i

and

2z_2 = 6 - 8i

. The sum of the new roots is

(6 + 8i) + (6 - 8i) = 12

, and the product is

(6 + 8i)(6 - 8i) = 36 - (8i)^2 = 36 + 64 = 100

. So the quadratic is

x^2 - 12x + 100 = 0

. Choice B has the wrong sign for the middle term. Choices C and D incorrectly use the original sum and product without scaling.

Question 12

If $\alpha$ and $\beta$ are the roots of $x^2 - 4x + 13 = 0$ , what is the value of $|\alpha - \beta|$ ?

$2\sqrt{3}$ using root difference formula
$4\sqrt{3}$ using root difference formula
$6$ using direct calculation methods (correct answer)
$2\sqrt{6}$ using discriminant relationships

Explanation: The roots are

\alpha = \frac{4 + \sqrt{16-52}}{2} = \frac{4 + \sqrt{-36}}{2} = 2 + 3i

and

\beta = 2 - 3i

. Therefore,

|\alpha - \beta| = |(2+3i) - (2-3i)| = |6i| = 6

. Alternatively, we can use the formula

|\alpha - \beta| = \frac{\sqrt{|\text{discriminant}|}}{|a|} = \frac{\sqrt{36}}{1} = 6

. Choices A, B, and D represent common errors in discriminant calculations or complex number arithmetic.

Question 13

The equation $4x^2 - 8x + 13 = 0$ has solutions $\alpha$ and $\beta$ . What is the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ ?

$\frac{13}{8}$
$-\frac{8}{13}$
$\frac{8}{13}$ (correct answer)
$-\frac{13}{8}$

Explanation: When you encounter a quadratic equation and need to find an expression involving the roots, Vieta's formulas are your most efficient tool. These formulas relate the coefficients of a polynomial to sums and products of its roots without requiring you to solve for the actual roots. For the quadratic

4x^2 - 8x + 13 = 0

with roots

\alpha

and

\beta

, Vieta's formulas tell us that

\alpha + \beta = \frac{8}{4} = 2

and

\alpha\beta = \frac{13}{4}

. To find

\frac{1}{\alpha} + \frac{1}{\beta}

, you need a common denominator:

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\alpha + \beta}{\alpha\beta}

. Substituting the values from Vieta's formulas:

\frac{2}{\frac{13}{4}} = 2 \times \frac{4}{13} = \frac{8}{13}

. Looking at the wrong answers: Choice A gives

\frac{13}{8}

, which results from incorrectly flipping the fraction

\frac{8}{13}

. Choice B shows

-\frac{8}{13}

, which happens if you mistakenly use

\alpha + \beta = -2

instead of

+2

(forgetting that for

ax^2 + bx + c = 0

, the sum is

-\frac{b}{a}

, and here

b = -8

). Choice D gives

-\frac{13}{8}

, combining both the sign error and the fraction flip. The key strategy here is recognizing that Vieta's formulas can solve root relationship problems without finding the actual roots. For any quadratic

ax^2 + bx + c = 0

, remember: sum of roots equals

-\frac{b}{a}

and product equals

\frac{c}{a}