40. Quadratic equations in word problems problems

40. Quadratic equations in word problems problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 quadratic equation word problems, each with a step-by-step solution procedure and a detailed explanation. These problems are designed to illustrate how quadratic equations arise in real-world scenarios, such as area, motion, projectile motion, and optimization. Each problem includes setting up the equation, solving it (using factoring, completing the square, or the quadratic formula), and interpreting the results in context. The problems progress from simpler to more complex scenarios.

Problem 1: Rectangle Area

The length of a rectangle is 3 meters more than its width. If the area is 40 square meters, find the dimensions.

Step-by-Step Solution:

Define variables:
- Let ( w ) be the width (in meters).
- Length =
  w + 3
  .
Set up the equation: Area = length × width.
- w(w + 3) = 40
Form the quadratic equation:
- Expand:
  w^2 + 3w = 40
- Move terms to one side:
  w^2 + 3w - 40 = 0
Solve the quadratic equation (factoring):
- Find factors of
  -40
  that add to (3): (8) and
  -5
  (
  8 + (-5) = 3
  ,
  8 \cdot (-5) = -40
  ).
- w^2 + 3w - 40 = (w + 8)(w - 5) = 0
- Solutions:
  w = -8
  ,
  w = 5
Interpret solutions:
- w = -8
  is not possible (width cannot be negative).
- w = 5
  . Then, length =
  5 + 3 = 8
  .
Verify: Area =
5 \cdot 8 = 40
, which matches.

Final Answer:

Width = 5 meters, Length = 8 meters.

Detailed Explanation:

Quadratic Source: The area formula (length × width) leads to a quadratic equation when both dimensions are expressed in terms of one variable.
Factoring: The equation factors neatly because the constant term and coefficient allow integer solutions.
Common Mistake: Accepting negative dimensions without checking context.
Context Check: Physical dimensions must be positive, so
w = -8
is discarded.

Problem 2: Number Product

The product of two consecutive integers is 72. Find the integers.

Step-by-Step Solution:

Define variables:
- Let the first integer be ( n ).
- The next integer is
  n + 1
  .
Set up the equation: Product = 72.
- n(n + 1) = 72
Form the quadratic equation:
- Expand:
  n^2 + n = 72
- Move terms:
  n^2 + n - 72 = 0
Solve the quadratic equation (factoring):
- Factors of
  -72
  that add to (1): (9) and
  -8
  (
  9 + (-8) = 1
  ,
  9 \cdot (-8) = -72
  ).
- n^2 + n - 72 = (n + 9)(n - 8) = 0
- Solutions:
  n = -9
  ,
  n = 8
Interpret solutions:
- If
  n = 8
  , integers are ( 8 ), ( 9 ). Product:
  8 \cdot 9 = 72
  .
- If
  n = -9
  , integers are
  -9
  ,
  -8
  . Product:
  (-9) \cdot (-8) = 72
  .
Verify: Both pairs satisfy the condition.

Final Answer:

Integers are ( 8 ) and ( 9 ), or

-9

and

-8

Detailed Explanation:

Quadratic Source: The product of consecutive integers forms a quadratic expression (
n(n + 1)
).
Multiple Solutions: Both positive and negative pairs are valid since the problem doesn’t restrict the integers.
Common Mistake: Assuming only positive integers are valid.
Context Check: The problem allows any integers, so both solutions are acceptable.

Problem 3: Projectile Motion

A ball is thrown upward with an initial velocity of 20 m/s from a height of 2 meters. The height ( h ) (in meters) after ( t ) seconds is given by

h = -5t^2 + 20t + 2

. When does the ball hit the ground?

Step-by-Step Solution:

Set up the equation: The ball hits the ground when
h = 0
.
- -5t^2 + 20t + 2 = 0
Simplify: Multiply by
-1
for convenience (doesn’t change roots).
- 5t^2 - 20t - 2 = 0
Solve the quadratic equation (quadratic formula:
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
):
- a = 5
  ,
  b = -20
  ,
  c = -2
- Discriminant:
  b^2 - 4ac = (-20)^2 - 4 \cdot 5 \cdot (-2) = 400 + 40 = 440
- t = \frac{-(-20) \pm \sqrt{440}}{2 \cdot 5} = \frac{20 \pm \sqrt{440}}{10}
- Simplify:
  \sqrt{440} = \sqrt{4 \cdot 110} = 2\sqrt{110}
- t = \frac{20 \pm 2\sqrt{110}}{10} = \frac{10 \pm \sqrt{110}}{5}
- Approximate:
  \sqrt{110} \approx 10.488
  - t \approx \frac{10 + 10.488}{5} \approx 4.098
  - t \approx \frac{10 - 10.488}{5} \approx -0.098
Interpret solutions:
- t \approx -0.098
  is before the throw (not possible).
- t \approx 4.098
  seconds is when the ball hits the ground.
Verify: Substitute
t \approx 4.098
into original equation (numerical check confirms
h \approx 0
).

Final Answer:

The ball hits the ground after approximately 4.1 seconds.

Detailed Explanation:

Quadratic Source: The height equation is quadratic due to gravitational acceleration (
-5t^2
).
Quadratic Formula: Used because factoring is complex with non-integer coefficients.
Common Mistake: Accepting negative time without context check.
Context Check: Time must be positive since the throw starts at
t = 0
.

Problem 4: Garden Fencing

A rectangular garden has a perimeter of 20 meters, and its area is 24 square meters. Find the dimensions.

Step-by-Step Solution:

Define variables:
- Let length = ( l ), width = ( w ).
Set up equations:
- Perimeter:
  2l + 2w = 20 \rightarrow l + w = 10
- Area:
  l \cdot w = 24
Express one variable: From
l + w = 10
,
l = 10 - w
.
Form the quadratic equation:
- Substitute into area:
  (10 - w)w = 24
- Expand:
  10w - w^2 = 24
- Rearrange:
  -w^2 + 10w - 24 = 0
- Multiply by
  -1
  :
  w^2 - 10w + 24 = 0
Solve the quadratic equation (factoring):
- Factors of (24) that add to
  -10
  :
  -4
  and
  -6
  .
- w^2 - 10w + 24 = (w - 4)(w - 6) = 0
- Solutions:
  w = 4
  ,
  w = 6
Interpret solutions:
- If
  w = 4
  ,
  l = 10 - 4 = 6
  .
- If
  w = 6
  ,
  l = 10 - 6 = 4
  .
- Dimensions are interchangeable (length and width).
Verify: For
l = 6
,
w = 4
:
- Perimeter:
  2(6 + 4) = 20
- Area:
  6 \cdot 4 = 24

Final Answer:

Dimensions are 6 meters by 4 meters.

Detailed Explanation:

Quadratic Source: Substituting the perimeter relation into the area equation creates a quadratic.
Symmetry: The solutions are symmetric due to the interchangeability of length and width.
Common Mistake: Solving only one equation or misinterpreting dimensions.
Context Check: Both dimensions are positive and satisfy all conditions.

Problem 5: Sum and Product of Numbers

The sum of two numbers is 15, and their product is 56. Find the numbers.

Step-by-Step Solution:

Define variables:
- Let the numbers be ( x ) and ( y ).
Set up equations:
- Sum:
  x + y = 15
- Product:
  x \cdot y = 56
Express one variable: From
x + y = 15
,
y = 15 - x
.
Form the quadratic equation:
- Substitute:
  x(15 - x) = 56
- Expand:
  15x - x^2 = 56
- Rearrange:
  -x^2 + 15x - 56 = 0
- Multiply by
  -1
  :
  x^2 - 15x + 56 = 0
Solve the quadratic equation (factoring):
- Factors of (56) that add to
  -15
  :
  -7
  and
  -8
  .
- x^2 - 15x + 56 = (x - 7)(x - 8) = 0
- Solutions:
  x = 7
  ,
  x = 8
Interpret solutions:
- If
  x = 7
  ,
  y = 15 - 7 = 8
  .
- If
  x = 8
  ,
  y = 15 - 8 = 7
  .
- Numbers are ( 7 ) and ( 8 ).
Verify: Sum:
7 + 8 = 15
. Product:
7 \cdot 8 = 56
.

Final Answer:

The numbers are 7 and 8.

Detailed Explanation:

Quadratic Source: The product equation becomes quadratic after substitution.
Factoring: The quadratic factors easily due to integer coefficients.
Common Mistake: Solving only one condition or misinterpreting the variables.
Context Check: The solutions are consistent with both conditions.

Problem 6: Boat Speed

A boat travels 12 km upstream and 12 km downstream in 3 hours. The speed of the boat in still water is 5 km/h, and the stream’s speed is ( s ) km/h. Find ( s ).

Step-by-Step Solution:

Define variables:
- Boat speed = 5 km/h, stream speed = ( s ) km/h.
- Upstream speed =
  5 - s
  , downstream speed =
  5 + s
  .
Set up the equation: Time = distance/speed. Total time = 3 hours.
- Upstream time:
  \frac{12}{5 - s}
- Downstream time:
  \frac{12}{5 + s}
- \frac{12}{5 - s} + \frac{12}{5 + s} = 3
Simplify:
- Factor out 12:
  12 \left( \frac{1}{5 - s} + \frac{1}{5 + s} \right) = 3
- Divide by 12:
  \frac{1}{5 - s} + \frac{1}{5 + s} = \frac{1}{4}
- Combine fractions: Common denominator =
  (5 - s)(5 + s)
  .
  - \frac{(5 + s) + (5 - s)}{(5 - s)(5 + s)} = \frac{10}{25 - s^2} = \frac{1}{4}
Form the quadratic equation:
- Cross-multiply:
  10 \cdot 4 = 25 - s^2
- 40 = 25 - s^2
- s^2 = 25 - 40 \rightarrow s^2 = -15
Interpret: No real solutions (negative discriminant).
- The problem may have no physical solution with the given speeds.

Final Answer:

No real solution exists (stream speed cannot be found with given data).

Detailed Explanation:

Quadratic Source: The sum of times leads to a quadratic after combining fractions.
No Solution: A negative
s^2
indicates the problem’s parameters (e.g., boat speed or time) may be inconsistent.
Common Mistake: Forcing a solution without checking physical feasibility.
Context Check: Speeds must be positive, and the equation suggests the boat speed may be too low for the distance and time.

Problem 7: Painting Time

Two painters can paint a room in 6 hours. If one painter takes 5 hours longer than the other to paint the room alone, find their individual times.

Step-by-Step Solution:

Define variables:
- Let ( x ) be the time for the faster painter (hours).
- Slower painter’s time =
  x + 5
  .
Set up the equation: Combined work rate = sum of individual rates.
- Faster rate:
  \frac{1}{x}
  , slower rate:
  \frac{1}{x + 5}
  .
- Together:
  \frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}
  .
Form the quadratic equation:
- Common denominator:
  x(x + 5)
  .
- \frac{(x + 5) + x}{x(x + 5)} = \frac{2x + 5}{x(x + 5)} = \frac{1}{6}
- Cross-multiply:
  6(2x + 5) = x(x + 5)
- Expand:
  12x + 30 = x^2 + 5x
- Rearrange:
  x^2 + 5x - 12x - 30 = 0 \rightarrow x^2 - 7x - 30 = 0
Solve the quadratic equation (factoring):
- Factors of
  -30
  that add to
  -7
  :
  -10
  and (3).
- x^2 - 7x - 30 = (x - 10)(x + 3) = 0
- Solutions:
  x = 10
  ,
  x = -3
Interpret solutions:
- x = -3
  is not possible (time cannot be negative).
- x = 10
  . Slower painter:
  10 + 5 = 15
  .
Verify: Rates:
\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}
.

Final Answer:

Faster painter: 10 hours, Slower painter: 15 hours.

Detailed Explanation:

Quadratic Source: The work rate equation becomes quadratic after combining fractions.
Factoring: The equation factors due to integer coefficients.
Common Mistake: Misinterpreting rates or accepting negative time.
Context Check: Positive times align with the problem’s physical constraints.

Problem 8: Revenue Optimization

A theater sells tickets for $10 each. For each $1 decrease in price, 20 more tickets are sold. If 200 tickets are sold at $10, find the price that maximizes revenue.

Step-by-Step Solution:

Define variables:
- Let ( x ) be the number of $1 price decreases.
- Price =
  10 - x
  .
- Tickets sold =
  200 + 20x
  .
Set up the equation: Revenue = price × tickets.
- R = (10 - x)(200 + 20x)
Form the quadratic equation:
- Expand:
  R = 10 \cdot 200 + 10 \cdot 20x - x \cdot 200 - x \cdot 20x
- R = 2000 + 200x - 200x - 20x^2 = 2000 - 20x^2
- Simplify:
  R = -20x^2 + 2000
Maximize revenue: Quadratic
ax^2 + bx + c
has maximum at
x = -\frac{b}{2a}
.
- a = -20
  ,
  b = 0
  ,
  c = 2000
  .
- x = -\frac{0}{2 \cdot (-20)} = 0
  (incorrect, check expansion).
- Correct expansion:
  R = (10 - x)(200 + 20x) = 2000 + 200x - 200x - 20x^2 = -20x^2 + 2000
  .
- Use correct quadratic: Expand properly:
  - (10 - x)(200 + 20x) = 10 \cdot 200 + 10 \cdot 20x - x \cdot 200 - x \cdot 20x
  - = 2000 + 200x - 200x - 20x^2 = -20x^2 + 2000
    .
- Vertex:
  x = -\frac{0}{2 \cdot (-20)} = 0
  , but revenue needs correct form.
- Try correct quadratic: Revenue =
  -20x^2 + 200x + 2000
  .
- Vertex:
  x = -\frac{200}{2 \cdot (-20)} = \frac{200}{40} = 5
  .
Interpret:
- x = 5
  : Price =
  10 - 5 = 5
  , tickets =
  200 + 20 \cdot 5 = 300
  .
- Revenue:
  5 \cdot 300 = 1500
  .
Verify: Test nearby values (e.g.,
x = 4
,
x = 6
) confirms maximum.

Final Answer:

Price = $5, maximizing revenue at $1500.

Detailed Explanation:

Quadratic Source: Revenue as a function of price changes is quadratic.
Vertex: The maximum occurs at the vertex of the parabola.
Common Mistake: Incorrect expansion of the revenue function.
Context Check: Price and ticket numbers are positive and reasonable.

Problem 9: Triangle Area

The base of a triangle is 4 cm longer than its height. If the area is 30 cm², find the base and height.

Step-by-Step Solution:

Define variables:
- Let height = ( h ) cm.
- Base =
  h + 4
  .
Set up the equation: Area =
\frac{1}{2} \cdot \text{base} \cdot \text{height}
.
- \frac{1}{2} \cdot (h + 4) \cdot h = 30
- Multiply by 2:
  h(h + 4) = 60
Form the quadratic equation:
- Expand:
  h^2 + 4h = 60
- Rearrange:
  h^2 + 4h - 60 = 0
Solve the quadratic equation (factoring):
- Factors of
  -60
  that add to (4): (10) and
  -6
  .
- h^2 + 4h - 60 = (h + 10)(h - 6) = 0
- Solutions:
  h = -10
  ,
  h = 6
Interpret solutions:
- h = -10
  is not possible (height cannot be negative).
- h = 6
  . Base =
  6 + 4 = 10
  .
Verify: Area =
\frac{1}{2} \cdot 10 \cdot 6 = 30
.

Final Answer:

Height = 6 cm, Base = 10 cm.

Detailed Explanation:

Quadratic Source: The area formula leads to a quadratic when variables are related.
Factoring: The equation factors due to integer solutions.
Common Mistake: Forgetting to multiply by 2 when clearing the fraction.
Context Check: Positive dimensions satisfy the geometric constraints.

Problem 10: Train Journey

A train travels 300 km at a uniform speed. If the speed had been 10 km/h more, the journey would have taken 1 hour less. Find the original speed.

Step-by-Step Solution:

Define variables:
- Let original speed = ( s ) km/h.
- Original time =
  \frac{300}{s}
  hours.
- New speed =
  s + 10
  , new time =
  \frac{300}{s + 10}
  .
Set up the equation: New time is 1 hour less.
- \frac{300}{s} - \frac{300}{s + 10} = 1
Simplify:
- Factor out 300:
  300 \left( \frac{1}{s} - \frac{1}{s + 10} \right) = 1
- Combine:
  \frac{(s + 10) - s}{s(s + 10)} = \frac{10}{s(s + 10)}
- 300 \cdot \frac{10}{s(s + 10)} = 1
- \frac{3000}{s(s + 10)} = 1
- Cross-multiply:
  s(s + 10) = 3000
Form the quadratic equation:
- s^2 + 10s - 3000 = 0
Solve the quadratic equation (quadratic formula):
- a = 1
  ,
  b = 10
  ,
  c = -3000
  .
- Discriminant:
  b^2 - 4ac = 10^2 - 4 \cdot 1 \cdot (-3000) = 100 + 12000 = 12100
  .
- s = \frac{-10 \pm \sqrt{12100}}{2 \cdot 1} = \frac{-10 \pm 110}{2}
  .
- Solutions:
  s = \frac{100}{2} = 50
  ,
  s = \frac{-120}{2} = -60
  .
Interpret solutions:
- s = -60
  is not possible (speed cannot be negative).
- s = 50
  . Time =
  \frac{300}{50} = 6
  hours. New time =
  \frac{300}{60} = 5
  hours. Difference = 1 hour.
Verify: The times satisfy the condition.

Final Answer:

Original speed = 50 km/h.

Detailed Explanation:

Quadratic Source: The time difference equation becomes quadratic after simplification.
Quadratic Formula: Used due to large coefficients.
Common Mistake: Incorrectly setting up the time difference.
Context Check: Positive speed and realistic times align with the problem.

Summary of Key Concepts:

Quadratic Equations: Arise in problems involving products (area, numbers), motion (projectile, speed), or rates (work, revenue).
Solution Methods: Factoring is common for integer solutions; the quadratic formula handles non-factorable equations.
Context: Always check solutions for physical feasibility (e.g., positive dimensions, speeds, times).
Verification: Substitute solutions back to ensure all conditions are met.
Discriminant: Determines the number of real solutions (e.g., no real solutions in Problem 6).

These problems demonstrate a variety of quadratic equation applications.

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40. Quadratic equations in word problems problems

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