Algebra Practice

Quadratic Word Problems Practice Test

Translate real situations into quadratic equations, solve them, and reject infeasible roots.

Quadratic Word Problems Practice Test

This test has 20 questions

Instant feedback · Worked explanations

After the test · quadratic modeling field guide

Quadratic word problems are solved twice: first algebraically, then in the real situation

This practice set emphasizes translation from words into quadratic equations. The contexts include area, consecutive integers, projectile motion, revenue, fencing, right triangles, and square-root models. Solving the equation may produce two algebraic roots, but the work is not finished until dimensions, time, and other physical restrictions are checked and the exact requested quantity is reported.

areaconsecutive integersprojectile motion revenuefencingright triangles square-root modelsfeasibility checks
A mathematically correct root can still be the wrong answer to a word problem if it represents a negative length, negative time, or the wrong requested quantity.

Story-to-answer pipeline

Keep modeling, solving, and interpretation as separate stages

Define State what x represents.

Include units so later roots can be interpreted correctly.

Translate Build the quadratic model from the relationships in the story.

Dimensions, rates, areas, revenue, or geometric constraints create the equation.

Solve Use factoring, square roots, or another appropriate quadratic method.

If factoring, first move all terms to one side.

Filter Reject roots that violate the context.

Negative lengths and negative physical times are typical infeasible values.

Report Answer the exact quantity requested.

If x is only a parameter, convert it to x − 2, x + 5, or another requested dimension before finishing.

Quick application formula shelf

Useful relationships for the contexts named on the page

Rectangle / fencing Area and perimeter A = LW
P = 2L + 2W

If one dimension is written in terms of x, substitution often creates the quadratic.

Right triangle Pythagorean relationship a² + b² = c²

When side lengths are linear expressions in x, squaring them creates a quadratic equation.

Projectile motion Quadratic height model h(t) = at² + bt + c

Zeros represent ground-level times; the vertex represents maximum or minimum height.

Revenue Price times quantity R = price × quantity

If price or demand is linear in x, their product can create a quadratic revenue function.

Consecutive integers Write neighboring values from one variable x, x + 1
or x, x + 2

The exact expressions depend on whether the problem uses consecutive integers or consecutive even/odd integers.

Vertex optimization Extreme point of ax² + bx + c x = −b / (2a)

The vertex answers maximum or minimum questions, but the requested output may be the y-value rather than the x-value.

Context atlas

Look for the relationship that naturally creates a product or a square

Area & fencing

Express both dimensions in one variable, then multiply for area or combine them in the perimeter condition.

L(x) · W(x) = area

Projectile motion

The quadratic model gives height over time. Solve h(t)=0 for ground contact or use the vertex for peak height.

h(t) = at² + bt + c

Revenue

A changing price multiplied by a changing quantity often produces a quadratic revenue model.

R(x) = p(x)q(x)

Right triangles

Insert side expressions into the Pythagorean theorem, expand, and solve the resulting quadratic.

a(x)² + b(x)² = c(x)²

Consecutive integers

Represent all integers from one variable, then translate a product or sum condition.

x(x + 1), x(x + 2), …

Square-root models

After isolating a squared expression, use both square-root branches and then apply contextual restrictions.

u² = k → u = ±√k

Visual model · dimensions and area

The variable may be only a building block for the actual dimensions

x − 2
x + 5
A = (x − 2)(x + 5)
Model

If the area is known, set (x − 2)(x + 5) equal to that area and solve the quadratic.

Feasibility

A root must make both dimensions positive if they represent physical lengths.

Requested quantity

If the question asks for the shorter side, report x − 2 — not automatically the value of x.

Visual model · projectile motion

Zeros and the vertex answer different questions in a motion model

Ground contact

Solve h(t)=0. A negative algebraic time may exist but is usually infeasible for elapsed time after launch.

Maximum height

Use the vertex. The vertex time and vertex height are different quantities.

Interpret units

Report seconds for time and the model's distance unit for height.

Feasibility gate

Algebra produces candidate roots; context decides which candidates survive

This is the central interpretation skill described on the page.

Candidate passes

x = 8
dimension = x − 2 = 6

The resulting length is positive, so it can represent a physical dimension.

CHECK
UNITS +
RESTRICTIONS

Candidate rejected

x = −3
time = −3 s

If the model represents elapsed time after launch, a negative value is not feasible in that context.

Answer what was actually asked

The solved parameter is not always the final quantity

Algebraic result

x = 7

This is the model parameter.

QUESTION
ASKS FOR
x − 2

Contextual answer

x − 2 = 5

If x − 2 is the requested dimension, 5 with the correct units is the final answer.

Vertex optimization

Maximum and minimum questions are vertex questions

Identify the quadratic

f(x) = ax² + bx + c

Revenue, area under a constraint, or projectile height can all lead to a quadratic objective.

Find vertex input

xᵥ = −b / (2a)

This is the x-value where the maximum or minimum occurs.

Report the requested output

yᵥ = f(xᵥ)

If the question asks for maximum revenue or maximum height, evaluate the function at the vertex input.

x
x + 2
x + 4

Visual model · right triangles

Linear side expressions become quadratic after squaring

A right-triangle application can translate directly through the Pythagorean theorem. After solving for x, check that every resulting side length is positive and that the requested side is evaluated from its own expression.

x² + (x + 2)² = (x + 4)²

Solve the model efficiently

The story creates the quadratic; its algebraic form suggests the solving method

Factoring
Use when the model reduces to a factorable quadratic in zero form.
Move every term to one side first, then apply the zero-product property.
Square roots
Use when a squared expression can be isolated directly.
Keep both ± branches algebraically, then reject any infeasible contextual result.
Vertex
Use for maximum or minimum questions.
Distinguish the vertex input from the maximum/minimum output.

Common mistakes from the page

The most important errors happen after or before the algebra, not inside it

Variable not defined
Writing an equation before stating what x represents. Define x with units first.

This makes it possible to interpret every root correctly later.

Factoring before zero form
Trying to use the zero-product property while the equation is not equal to zero. Move all terms to one side first.

Factoring solves a quadratic through a product equal to zero.

Negative length or time kept
Reporting every algebraic root without checking physical meaning. Reject values that violate the model's real-world restrictions.

Negative dimensions and negative elapsed times are typical examples.

x reported instead of x − 2
The equation gives x = 7, so the final answer is stated as 7 even though the requested side is x − 2. Evaluate the requested expression: x − 2 = 5.

Always return from the algebra to the original variable definition and question wording.

Vertex coordinate confused
Giving the vertex x-coordinate when the question asks for maximum revenue or height. Use xᵥ to locate the optimum, then calculate f(xᵥ) when the output is requested.

The input and the maximum/minimum value are different quantities.

Final word-problem checklist

A correct contextual answer must survive five checks

Variable defined x has a clear meaning and unit before the quadratic model is built.
Model translated correctly The area, motion, revenue, geometry, or other relationship matches the story.
Roots filtered Negative or otherwise impossible dimensions and times are rejected when context requires it.
Requested quantity reported The final answer gives the actual dimension, time, revenue, height, or other requested quantity with units.