Catapult KitArcfolk · quadratics

Lobs a ball in a perfect arc: projectile and area problems.

Quadratic Word Problems

Projectile and area problems. In MathBound, the free Algebra I adventure game, Catapult Kit asks these problems, and the only way past is to solve them.

Part of Solving Quadratic Equations: chapter 10 of the standard course.

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10 questions, no sign-up, nothing saved. Climbs from the basics to harder questions.

What you will practice

Consecutive Integers. Translate and solve a word problem involving the product of consecutive integers yielding a quadratic equation.
Geometric Area. Solve an area application for a rectangle where one dimension is defined in terms of the other (A=lw).
Pythagorean Theorem. Use the Pythagorean theorem to solve for missing sides of a right triangle where the side lengths are given as algebraic expressions.
Projectile Motion. Solve a given quadratic projectile motion equation for the time it takes an object to hit the ground (h=0).

Worked examples

Made by the game’s own question engine. Try each one, then open its steps.

Example 1 · Consecutive Integers

The

product

of

two

consecutive

odd

positive

integers

is

63 .

Find

them.

$7$ $and$ $8$
$9$ $and$ $11$
$7$ $and$ $9$
$5$ $and$ $7$

Show the worked steps

$Let$ $the$ $integers$ $be$ $n$ $and$ $n + 2,$ $so$ $n^{2} + 2 n = 63 .$
$n^{2} + 2 n - 63 = 0$ $gives$ $n = 7 (the$ $other$ $root,$ $- 9,$ $is$ $negative) .$
$The$ $integers$ $are$ $7$ $and$ $9 .$

Answer: $7$ $and$ $9$

Example 2 · Geometric Area

Each

side

of

a

square

patio

is

lengthened

by

2

meters,

giving

an

area

of

81

square

meters.

How

long

was

each

side

at

first?

$11$ $meters$
$9$ $meters$
$7$ $meters$
$20$ $meters$

Show the worked steps

$The$ $new$ $side$ $is$ $s + 2,$ $so$ $(s + 2)^{2} = 81 .$
$s + 2 = 9,$ $the$ $positive$ $root,$ $so$ $s = 7$ $meters.$

Answer: $7$ $meters$

Example 3 · Pythagorean Theorem

One

leg

of

a

right

triangle

is

7

units

shorter

than

the

other.

The

hypotenuse

is

17

units.

Find

the

longer

leg.

$8$ $units$
$10$ $units$
$24$ $units$
$15$ $units$

Show the worked steps

$Let$ $the$ $longer$ $leg$ $be$ $x,$ $so$ $the$ $other$ $is$ $x - 7 .$
$x^{2} + (x - 7)^{2} = 289$ $gives$ $2 x^{2} - 14 x - 240 = 0.$
$The$ $positive$ $root$ $is$ $x = 15 .$

Answer: $15$ $units$

Where it is in your textbook

Textbook	Lessons
Standard course	10.4
Big Ideas Math Algebra 1	8.2, 9.5
Big Ideas Math Algebra 1 (2022)	8.2, 9.5
Carnegie Learning Algebra I	M5 1.1, M5 3.3
CPM Core Connections Algebra	9.1.4, 10.2.6
CPS Skyline Algebra	7.12-7.13
enVision Algebra 1	8-4
enVision Algebra 1 (2024)	8-4
Eureka Math Algebra I (EngageNY)	4.6-4.7, 4.23-4.24, 5.8-5.9
Glencoe Algebra 1	9-5, 10-5
HMH Into Algebra 1	18.4, 19.5
Holt McDougal Larson Algebra 1	9.6
Illustrative Mathematics Algebra 1	6.5-6.7, 7.1-7.2, 7.16-7.19, 7.24
Illustrative Mathematics Algebra 1 (v.360)	7.5-7.7, 8.1-8.2, 8.16-8.19, 8.24
Math Nation Algebra 1 (Florida B.E.S.T.)	8.8, 9.4, 9.11
Open Up NC Math 1 (MVP)	6.4
OpenStax Elementary Algebra 2e	10.4
Pearson Algebra 1 Common Core	9-6, 10-1
Reveal Algebra 1	11-6
SpringBoard Algebra 1	33

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