Complex Numbers and the Quadratic Formula: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Simplify $i^{42}$ .

Answer choices for question 1 A $42i$ B $1$ C $-1$ D $-i$
2

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Simplify $(9 - 4i) - (5 + 3i)$ .

Answer choices for question 2 A $4 - 7i$ B $4 - i$ C $-3$ D $-4 + 7i$
3

What number $k$ makes $x^2 - 14x + k$ a perfect square?

Answer choices for question 3 A $-7$ B $196$ C $-49$ D $49$
4

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Simplify $\sqrt{-48}$ .

Answer choices for question 4 A $-4\sqrt{3}$ B $4i\sqrt{3}$ C $16i\sqrt{3}$ D $4\sqrt{3}$
5

When $2x^2 - 5x = 3 - x$ is written in the standard form $ax^2 + bx + c = 0$ , what are $a$ , $b$ , and $c$ ?

Answer choices for question 5 A $a = 2$ , $b = -4$ , $c = -3$ B $a = 2$ , $b = -5$ , $c = 3$ C $a = 2$ , $b = -4$ , $c = 3$ D $a = 2$ , $b = -6$ , $c = -3$
6

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Write $(4 + 3i)(2 - 5i)$ in standard form $a + bi$ .

Answer choices for question 6 A $-7 - 14i$ B $8 - 14i$ C $23 - 14i$ D $-21$
7

What kind of roots does the equation $2x^2 + 4x = -8$ have?

Answer choices for question 7 A two distinct real roots B two complex conjugate roots C one repeated real root D no roots at all
8

Advanced. This question goes beyond core Algebra I. It is not required by the course.

What is the product of $-3 + 5i$ and its conjugate?

Answer choices for question 8 A $-16$ B $-34$ C $16 + 30i$ D $34$
9

Solve $3x^2 + 12x + 27 = 0$ .

Answer choices for question 9 A $x = -2 \pm i\sqrt{5}$ B $x = -3$ or $x = -9$ C $x = -2 \pm 5i$ D $x = -2 \pm \sqrt{7}$
10

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Write $\left(4 + 3i^{9}\right)\left(2 - i^{11}\right)$ in standard form $a + bi$ .

Answer choices for question 10 A $305 + 10i$ B $11 + 10i$ C $5 + 10i$ D $11 + 2i$
11

The square of a positive number is $30$ more than the number itself. What is the number?

Answer choices for question 11 A $-5$ B $5$ C There is no such number. D $6$
12

Solve $x^2 - 4x + 8 = 0$ .

Answer choices for question 12 A $x = -2 \pm 2i$ B $x = 2 \pm 2i$ C $x = 2 \pm 2\sqrt{3}$ D $x = 2 \pm 8i$
13

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Evaluate $\sqrt{-3}\cdot\sqrt{-27}$ .

Answer choices for question 13 A $9$ B $9i$ C $-9$ D $-\sqrt{30}$
14

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Write $\dfrac{3 - 2i}{4 + i}$ in standard form $a + bi$ .

Answer choices for question 14 A $\dfrac{10}{17} - \dfrac{11}{17}i$ B $\dfrac{14}{15} - \dfrac{11}{15}i$ C $\dfrac{14 - 5i}{15 + 8i}$ D $\dfrac{10}{17} - 11i$
15

Solve $x^2 + 7x + 13 = 0$ .

Answer choices for question 15 A $x = \dfrac{7 \pm i\sqrt{3}}{2}$ B $x = \dfrac{-7 \pm i\sqrt{3}}{2}$ C $x = \dfrac{-7 \pm \sqrt{3}}{2}$ D $x = -7 \pm \dfrac{i\sqrt{3}}{2}$
16

A stone is thrown upward from the edge of a wall $24$ feet high with an initial upward speed of $40$ feet per second, so its height in feet after $t$ seconds is $h = -16t^2 + 40t + 24$ . At what times is the stone $48$ feet above the ground?

Answer choices for question 16 A Only at $t = 3$ seconds. B Only at $t = \tfrac{3}{2}$ seconds. C The stone never reaches $48$ feet. D At $t = 1$ second and again at $t = \tfrac{3}{2}$ seconds.
17

Advanced. This question goes beyond core Algebra I. It is not required by the course.

If $(3 + 2i)w = 13$ , what is $w$ ?

Answer choices for question 17 A $3 - 2i$ B $1$ C $\dfrac{39}{5} - \dfrac{26}{5}i$ D $\dfrac{3}{13} - \dfrac{2}{13}i$
18

Solve $3x^2 - 5x - 4 = 0$ .

Answer choices for question 18 A $x = \dfrac{-5 \pm \sqrt{73}}{6}$ B $x = \dfrac{5 \pm \sqrt{23}}{6}$ C $x = \dfrac{5 \pm \sqrt{73}}{6}$ D $x = \dfrac{5 \pm \sqrt{73}}{3}$
19

For what value of $k$ does $2x^2 + 12x + k = 0$ have exactly one real root?

Answer choices for question 19 A $k = 36$ B $k = 6$ C $k = 9$ D $k = 18$
20

A rectangular deck is $2$ meters longer than twice its width, and its area is $84$ square meters. How wide is the deck?

Answer choices for question 20 A $-7$ meters B $6$ meters C $\left(-1 + \sqrt{85}\right)$ meters D $14$ meters

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Where you stand

Treat $i$ as the number with $i^2 = -1$ , and reduce a power of $i$ using the remainder of its exponent on division by $4$ . Review: Imaginary Numbers
Rewrite the square root of a negative number as $i$ times a real square root, and multiply such roots correctly. Review: Imaginary Numbers
Add, subtract and multiply complex numbers, reporting the result in the form $a + bi$ . Review: Arithmetic with Complex Numbers
Use the conjugate: that $(a+bi)(a-bi)$ is the real number $a^2+b^2$ , and that it clears $i$ from a denominator. Review: Arithmetic with Complex Numbers
Complete the square to rewrite a quadratic as a squared binomial plus a constant, and to solve one. Review: Completing the Square
Read $a$ , $b$ and $c$ from a quadratic in standard form and apply the quadratic formula. Review: The Quadratic Formula
Use the discriminant $b^2 - 4ac$ to decide whether the roots are two real, one repeated, or a complex conjugate pair. Review: The Quadratic Formula , Completing the Square
Model a situation with a quadratic, then decide which of its roots the situation actually permits. Review: Applications of Quadratics

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A weighted total

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Write $i^{22}(2-i)+(1+3i)$ in the form $a+bi$ .
1. Hint 1
  The power of $i$ is determined by its exponent remainder.
2. Hint 2
  Apply that value to the entire parenthesized number before combining the two parts.
Answer

$-1+4i$ .

Full solution

The exponent $22$ leaves remainder $2$ after division by $4$ , so $i^{22}=-1$ .
$i^{22}(2-i)=-2+i$

Add the remaining number.
$(-2+i)+(1+3i)=-1+4i$

Answer

$-1+4i$ .

Key idea

A power of $i$ acts on both parts of the complex number it multiplies.
Problem 1: Mark complete
Problem 2 Two recorded values

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A complex number $z$ has $z-\overline z=4i$ and $z\overline z=20$ , where $\overline z$ is its conjugate. Find $z$ if its real part is negative.
1. Hint 1
  Write $z=a+bi$ and express the two records using real $a$ and $b$ .
2. Hint 2
  The difference determines the imaginary part; then the conjugate product determines the square of the real part.
Answer

$z=-4+2i$ .

Full solution

Write $z=a+bi$ with $a$ and $b$ real, so
$\overline z=a-bi.$
The difference from the conjugate is $2bi$ , so $b=2$ .
Since $z\overline z=a^2+b^2$ , the product condition gives
$a^2+4=20$

Thus $a^2=16$ , and the negative real part selects $a=-4$ .
Check: $-4+2i$ and $-4-2i$ differ by $4i$ and multiply to $20$ .

Answer

$z=-4+2i$ .

Key idea

A conjugate difference and conjugate product can recover both parts when the remaining sign is specified.
Problem 2: Mark complete
Problem 3 An output expression

Write $(x+2)(x-6)+9$ in the form $(x-h)^2+k$ , where $h$ and $k$ are constants.
1. Hint 1
  Expand the product and collect the constant terms.
2. Hint 2
  Use half the linear coefficient inside the squared binomial and balance the constant outside it.
Answer

$(x-2)^2-7$ .

Full solution

Expansion gives
$(x+2)(x-6)+9=x^2-4x-3$

Half the linear coefficient is $-2$ , whose square is $4$ .
Add and subtract $4$ .
$x^2-4x-3=(x-2)^2-7$

Expanding the result returns $x^2-4x-3$ , confirming the identity.

Answer

$(x-2)^2-7$ .

Key idea

A product expression can be rewritten as one squared binomial plus a constant.
Problem 3: Mark complete
Problem 4 A ratio of records

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Let $z=4+3i$ and $w=1-i$ . Write $(z-w)/(z+w)$ in standard form $a+bi$ .
1. Hint 1
  Find the sum and difference before treating the quotient.
2. Hint 2
  The denominator is nonzero; multiply numerator and denominator by its conjugate.
Answer

$\dfrac{23}{29}+\dfrac{14}{29}i$ .

Full solution

The difference is $3+4i$ and the sum is $5+2i$ , which is nonzero.
Its conjugate is $5-2i$ , and multiplying by it gives the real denominator $5^2+2^2=29$ .
$\frac{z-w}{z+w}=\frac{(3+4i)(5-2i)}{29}$

Expanding the numerator, $(3+4i)(5-2i)$ gives $15-6i+20i-8i^2$ .
Since $i^2=-1$ , the last term is $+8$ , so the numerator is $23+14i$ .
The result is $23/29+(14/29)i$ , and multiplying it by $5+2i$ returns $3+4i$ .

Answer

$\dfrac{23}{29}+\dfrac{14}{29}i$ .

Key idea

Form the requested complex combinations before using a conjugate to divide them.
Problem 4: Mark complete
Problem 5 A stopping model

In a simplified vehicle model, a speed of $v$ feet per second gives a reaction distance of $3v$ feet and a braking distance of $v^2/4$ feet. Find the positive speed that gives a total stopping distance of $14$ feet, exactly.
1. Hint 1
  The total distance is the sum of the two modeled distances.
2. Hint 2
  Clear the fraction and solve the resulting quadratic; the speed must be positive.
Answer

$v=-6+2\sqrt{23}$ feet per second.

Full solution

The distance condition is
$3v+\frac{v^2}{4}=14$

Multiply by $4$ and gather terms.
$v^2+12v-56=0$

With coefficients $1$ , $12$ , and $-56$ , the formula gives
$v=\frac{-12\pm\sqrt{368}}2$

Since $368=16\times23$ , the radical simplifies to
$\sqrt{368}=4\sqrt{23}.$
Dividing both numerator terms by $2$ gives the roots $-6\pm2\sqrt{23}$ .
Since $\sqrt{23}>3$ , the plus choice is positive and the minus choice negative.
For $v=-6+2\sqrt{23}$ , squaring gives $v^2=128-24\sqrt{23}$ , and $12v=-72+24\sqrt{23}$ , so
$v^2+12v=56.$
Dividing by $4$ returns $3v+v^2/4=14$ , the total stopping distance in feet.

Answer

$v=-6+2\sqrt{23}$ feet per second.

Key idea

A modeled speed must satisfy the distance equation and the positive-speed condition.
Problem 5: Mark complete
Problem 6 A storage format

A program stores $R(x)=2x^2+10x+15$ in the form $2(x-h)^2+k$ . Find $h$ and $k$ , and determine the type of roots of $R(x)=0$ .
1. Hint 1
  The stored form has the same leading coefficient as $R(x)$ , so compare the two expressions term by term.
2. Hint 2
  The discriminant independently determines whether the zero-output inputs are real or nonreal.
Answer

$h=-\dfrac52$ ; $k=\dfrac52$ ; two nonreal complex conjugate roots.

Full solution

The target form expands to
$2x^2-4hx+2h^2+k$

Matching the linear coefficients gives $-4h=10$ , so $h=-5/2$ .
Matching the constants gives $2h^2+k=15$ , and $2h^2=25/2$ , so $k=5/2$ .
$R(x)=2\left(x+\frac52\right)^2+\frac52$

To solve $R(x)=0$ , divide every term by $2$ .
$x^2+5x+\frac{15}2=0$
Move the constant across and add $\left(\frac52\right)^2=\frac{25}4$ to both sides.
$\left(x+\frac52\right)^2=-\frac54$
The right side is negative, so the equation has no real root.

For the zero-output equation the discriminant is
$D=10^2-4(2)(15)$

This is $-20$ , so the roots are a nonreal complex conjugate pair.
The positive constant outside the positive square confirms there is no real zero.

Answer

$h=-\dfrac52$ ; $k=\dfrac52$ ; two nonreal complex conjugate roots.

Key idea

The constant outside a completed square and the discriminant give consistent information about real zeros.
Problem 6: Mark complete
Problem 7 An expanded display

A rectangular display has $r$ rows and $r+2$ columns of lights. Adding one full row and one full column creates a new display with $80$ lights. Find the original numbers of rows and columns, where both are positive whole numbers.
1. Hint 1
  Write the dimensions after both additions in terms of the original row count.
2. Hint 2
  Solve the new area equation and interpret the roots as counts.
Answer

$7$ rows and $9$ columns.

Full solution

The new dimensions are $r+1$ and $r+3$ .
$(r+1)(r+3)=80$

Expand and gather terms.
$r^2+4r-77=0$

Factor.
$(r-7)(r+11)=0$

The roots are $7$ and $-11$ .
Only $7$ is a positive whole number, giving $7$ rows and $9$ columns originally.
The expanded display has $8$ rows and $10$ columns, totaling $80$ lights.

Answer

$7$ rows and $9$ columns.

Key idea

A count recovered from a quadratic keeps only the root the situation permits; a negative root is not a count.
Problem 7: Mark complete
Problem 8 Reversed coefficients

Let $a$ , $b$ , and $c$ be real, with $a\ne0$ and $c\ne0$ . A student claims that $ax^2+bx+c=0$ and $cx^2+bx+a=0$ have the same number of distinct real roots. Is the claim correct? Explain.
1. Hint 1
  Both equations are quadratic under the stated nonzero conditions.
2. Hint 2
  Compare the discriminants and recall what each sign says about the number of distinct real roots.
Answer

Yes, the claim is correct.

Full solution

The first discriminant is
$D_1=b^2-4ac$

The second is
$D_2=b^2-4ca$

Because $ac=ca$ , the discriminants agree.
If the common value is positive both equations have two distinct real roots; if zero each has one; if negative neither has real roots.

Answer

Yes, the claim is correct.

Key idea

Interchanging nonzero leading and constant coefficients preserves the discriminant and hence the number of distinct real roots.
Problem 8: Mark complete
Problem 9 An inside walkway

A rectangular courtyard measures $20$ meters by $16$ meters. A walkway of uniform width $x$ meters runs along the inside of all four edges, leaving a rectangular lawn of area $96$ square meters. Find $x$ , and show that only one solution of the area equation describes a possible walkway.
1. Hint 1
  Each side of the lawn is shorter than the courtyard side it lies along by the walkway width at both ends.
2. Hint 2
  Solve the area equation, then test each solution against the shorter side of the courtyard.
Answer

$x=4$ meters; the other solution of the area equation, $x=14$ meters, is rejected.

Full solution

The lawn measures $20-2x$ by $16-2x$ meters.
$(20-2x)(16-2x)=96$

Expanding the left side gives $4x^2-72x+320$ .
Subtract $96$ from both sides.
$4x^2-72x+224=0$

Divide every term by $4$ .
$x^2-18x+56=0$

Move the constant across and add $81$ to both sides.
$(x-9)^2=25$

Thus $x-9=\pm5$ , giving $x=4$ or $x=14$ .
The two strips along the longer edges together take $2x$ meters of the $16$ -meter side, so $x=14$ would need $28$ meters there and is impossible, leaving $x=4$ meters.
The lawn is then $12$ meters by $8$ meters, an area of $96$ square meters.

Answer

$x=4$ meters; the other solution of the area equation, $x=14$ meters, is rejected.

Key idea

A root of an area equation still has to fit inside the figure it is meant to describe.
Problem 9: Mark complete
Problem 10 A coefficient choice

For real $m\ne1$ , consider $(m-1)x^2+2x+1=0$ . Classify the roots when $m=0$ , $m=2$ , and $m=3$ , and find all roots for $m=3$ .
1. Hint 1
  The leading coefficient changes with the chosen value of $m$ .
2. Hint 2
  Express the discriminant in terms of $m$ , then apply the formula for the requested choice.
Answer

$m=0$ : two distinct real roots. $m=2$ : one repeated real root. $m=3$ : nonreal conjugates $x=(-1\pm i)/2$ .

Full solution

The coefficients are $m-1$ , $2$ , and $1$ .
$D=4-4(m-1)$

This simplifies to $8-4m$ .
At $m=0$ it is positive, at $m=2$ it is zero, and at $m=3$ it is negative.
For $m=3$ , the equation has coefficients $2$ , $2$ , and $1$ .
$x=\frac{-2\pm\sqrt{-4}}4$

Therefore
$x=(-1\pm i)/2.$
Substituting either root gives
$2x^2+2x+1=0.$

Answer

$m=0$ : two distinct real roots. $m=2$ : one repeated real root. $m=3$ : nonreal conjugates $x=(-1\pm i)/2$ .

Key idea

A parameter can move a quadratic between root types, so substitute it into every affected coefficient.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 10: Graphing Quadratics and Inequalities →

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