Quadratic Equation Word Problems Worksheets - Free Printable

Problem Sheet: Quadratic Equations

Here, I will solve each problem step by step.

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#### Problem 1:
Question: Abraham throws a ball from a point 40 m above the ground. The height of the ball from the ground level after \( t \) seconds is defined by the function \( h(t) = 40t - 5t^2 \). How long will the ball take to hit the ground?

Solution:
The ball hits the ground when its height \( h(t) \) is 0. Therefore, we need to solve the equation:
\[
h(t) = 40t - 5t^2 = 0
\]

1. Factor out the common term \( t \):
\[
t(40 - 5t) = 0
\]

2. Set each factor equal to zero:
\[
t = 0 \quad \text{or} \quad 40 - 5t = 0
\]

3. Solve for \( t \) in the second equation:
\[
40 - 5t = 0 \implies 5t = 40 \implies t = 8
\]

4. Interpret the solutions:
- \( t = 0 \) corresponds to the initial time when the ball is thrown.
- \( t = 8 \) corresponds to the time when the ball hits the ground.

Thus, the ball takes 8 seconds to hit the ground.

Answer:
\[
\boxed{8}
\]

---

#### Problem 2:
Question: The area of a rectangular pool is 1260 ft². Find the dimensions of the rectangle if one side of the pool is 48 ft more than three times the other side.

Solution:
Let the length of the shorter side be \( x \) ft. Then, the length of the longer side is \( 3x + 48 \) ft. The area of the rectangle is given by:
\[
\text{Area} = \text{length} \times \text{width} = x(3x + 48)
\]

We know the area is 1260 ft², so:
\[
x(3x + 48) = 1260
\]

1. Expand and simplify the equation:
\[
3x^2 + 48x = 1260
\]

2. Rearrange into standard quadratic form:
\[
3x^2 + 48x - 1260 = 0
\]

3. Divide the entire equation by 3 to simplify:
\[
x^2 + 16x - 420 = 0
\]

4. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 16 \), and \( c = -420 \):
\[
x = \frac{-16 \pm \sqrt{16^2 - 4(1)(-420)}}{2(1)}
\]
\[
x = \frac{-16 \pm \sqrt{256 + 1680}}{2}
\]
\[
x = \frac{-16 \pm \sqrt{1936}}{2}
\]
\[
x = \frac{-16 \pm 44}{2}
\]

5. Calculate the two possible values for \( x \):
\[
x = \frac{-16 + 44}{2} = \frac{28}{2} = 14
\]
\[
x = \frac{-16 - 44}{2} = \frac{-60}{2} = -30
\]

6. Since \( x \) represents a length, it must be positive. Thus, \( x = 14 \).

7. Find the longer side:
\[
3x + 48 = 3(14) + 48 = 42 + 48 = 90
\]

Thus, the dimensions of the rectangle are 14 ft and 90 ft.

Answer:
\[
\boxed{14 \text{ ft}, 90 \text{ ft}}
\]

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#### Problem 3:
Question: The sum of the squares of two consecutive natural numbers is 313. Find the numbers.

Solution:
Let the two consecutive natural numbers be \( n \) and \( n+1 \). The sum of their squares is given by:
\[
n^2 + (n+1)^2 = 313
\]

1. Expand and simplify the equation:
\[
n^2 + (n^2 + 2n + 1) = 313
\]
\[
2n^2 + 2n + 1 = 313
\]

2. Rearrange into standard quadratic form:
\[
2n^2 + 2n + 1 - 313 = 0
\]
\[
2n^2 + 2n - 312 = 0
\]

3. Divide the entire equation by 2 to simplify:
\[
n^2 + n - 156 = 0
\]

4. Solve the quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -156 \):
\[
n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-156)}}{2(1)}
\]
\[
n = \frac{-1 \pm \sqrt{1 + 624}}{2}
\]
\[
n = \frac{-1 \pm \sqrt{625}}{2}
\]
\[
n = \frac{-1 \pm 25}{2}
\]

5. Calculate the two possible values for \( n \):
\[
n = \frac{-1 + 25}{2} = \frac{24}{2} = 12
\]
\[
n = \frac{-1 - 25}{2} = \frac{-26}{2} = -13
\]

6. Since \( n \) represents a natural number, it must be positive. Thus, \( n = 12 \).

7. The consecutive natural numbers are \( n = 12 \) and \( n+1 = 13 \).

Answer:
\[
\boxed{12, 13}
\]

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#### Problem 4:
Question: Two faucets can fill a tank in 1 hour and 20 minutes. The time taken by faucet A alone to fill the tank is 2 hours more than faucet B were to fill the same tank separately. How long does it take faucet A alone to fill the tank?

Solution:
Let the time taken by faucet B to fill the tank alone be \( t \) hours. Then, the time taken by faucet A to fill the tank alone is \( t + 2 \) hours.

1. The rate at which faucet B fills the tank is \( \frac{1}{t} \) (tank per hour).
2. The rate at which faucet A fills the tank is \( \frac{1}{t+2} \) (tank per hour).

When both faucets work together, they fill the tank in 1 hour and 20 minutes, which is \( 1 + \frac{20}{60} = \frac{4}{3} \) hours. Their combined rate is:
\[
\frac{1}{t} + \frac{1}{t+2} = \frac{3}{4}
\]

3. Solve the equation:
\[
\frac{1}{t} + \frac{1}{t+2} = \frac{3}{4}
\]

4. Find a common denominator and combine the fractions:
\[
\frac{(t+2) + t}{t(t+2)} = \frac{3}{4}
\]
\[
\frac{2t + 2}{t(t+2)} = \frac{3}{4}
\]

5. Cross-multiply to eliminate the fractions:
\[
4(2t + 2) = 3t(t + 2)
\]
\[
8t + 8 = 3t^2 + 6t
\]

6. Rearrange into standard quadratic form:
\[
3t^2 + 6t - 8t - 8 = 0
\]
\[
3t^2 - 2t - 8 = 0
\]

7. Solve the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -2 \), and \( c = -8 \):
\[
t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-8)}}{2(3)}
\]
\[
t = \frac{2 \pm \sqrt{4 + 96}}{6}
\]
\[
t = \frac{2 \pm \sqrt{100}}{6}
\]
\[
t = \frac{2 \pm 10}{6}
\]

8. Calculate the two possible values for \( t \):
\[
t = \frac{2 + 10}{6} = \frac{12}{6} = 2
\]
\[
t = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3}
\]

9. Since \( t \) represents time, it must be positive. Thus, \( t = 2 \).

10. The time taken by faucet A to fill the tank alone is:
\[
t + 2 = 2 + 2 = 4 \text{ hours}
\]

Answer:
\[
\boxed{4}
\]

---

#### Problem 5:
Question: If one side of a square is increased by 10 cm and another side is increased by 5 cm, a rectangle is formed with an area that measures three times the area of the square. Find the length of the side of the square.

Solution:
Let the side length of the square be \( x \) cm. The area of the square is:
\[
x^2 \text{ cm}^2
\]

When one side is increased by 10 cm and the other side is increased by 5 cm, the dimensions of the resulting rectangle are \( x + 10 \) cm and \( x + 5 \) cm. The area of the rectangle is:
\[
(x + 10)(x + 5) \text{ cm}^2
\]

We are told that the area of the rectangle is three times the area of the square:
\[
(x + 10)(x + 5) = 3x^2
\]

1. Expand and simplify the equation:
\[
x^2 + 5x + 10x + 50 = 3x^2
\]
\[
x^2 + 15x + 50 = 3x^2
\]

2. Rearrange into standard quadratic form:
\[
x^2 + 15x + 50 - 3x^2 = 0
\]
\[
-2x^2 + 15x + 50 = 0
\]

3. Multiply the entire equation by \(-1\) to simplify:
\[
2x^2 - 15x - 50 = 0
\]

4. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -15 \), and \( c = -50 \):
\[
x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(-50)}}{2(2)}
\]
\[
x = \frac{15 \pm \sqrt{225 + 400}}{4}
\]
\[
x = \frac{15 \pm \sqrt{625}}{4}
\]
\[
x = \frac{15 \pm 25}{4}
\]

5. Calculate the two possible values for \( x \):
\[
x = \frac{15 + 25}{4} = \frac{40}{4} = 10
\]
\[
x = \frac{15 - 25}{4} = \frac{-10}{4} = -2.5
\]

6. Since \( x \) represents the side length of a square, it must be positive. Thus, \( x = 10 \).

Answer:
\[
\boxed{10}
\]

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Final Answers:

1. \(\boxed{8}\)
2. \(\boxed{14 \text{ ft}, 90 \text{ ft}}\)
3. \(\boxed{12, 13}\)
4. \(\boxed{4}\)
5. \(\boxed{10}\)