Quadratic Equation Word Problems Worksheet

Let's solve each problem step by step.

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1) 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?

Step 1: Understand the equation
The height function is:
$$
h(t) = 40t - 5t^2
$$
We want to find when the ball hits the ground, i.e., when $ h(t) = 0 $.

Step 2: Set the equation to zero
$$
40t - 5t^2 = 0
$$

Step 3: Factor
$$
5t(8 - t) = 0
$$

Step 4: Solve for $ t $
$$
5t = 0 \Rightarrow t = 0 \\
8 - t = 0 \Rightarrow t = 8
$$

- $ t = 0 $ is the initial time (when the ball is thrown).
- $ t = 8 $ is when it hits the ground.

✔ Answer: The ball takes 8 seconds to hit the ground.

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2) 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.

Step 1: Let variables
Let the shorter side be $ x $ ft.
Then the longer side is $ 3x + 48 $ ft.

Step 2: Area formula
$$
\text{Area} = \text{length} \times \text{width} = x(3x + 48) = 1260
$$

Step 3: Expand and solve
$$
3x^2 + 48x = 1260 \\
3x^2 + 48x - 1260 = 0
$$

Divide entire equation by 3:
$$
x^2 + 16x - 420 = 0
$$

Step 4: Factor or use quadratic formula
Try factoring:
We need two numbers that multiply to $-420$ and add to $16$.

Try $ 30 $ and $-14$: $ 30 \times (-14) = -420 $, $ 30 + (-14) = 16 $ → yes!

$$
(x + 30)(x - 14) = 0
$$

So $ x = -30 $ or $ x = 14 $

Discard negative value: $ x = 14 $

Then the other side is:
$$
3(14) + 48 = 42 + 48 = 90
$$

✔ Answer: Dimensions are 14 ft and 90 ft.

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3) The sum of the squares of two consecutive natural numbers is 313. Find the numbers.

Step 1: Let the numbers be $ x $ and $ x+1 $

Sum of squares:
$$
x^2 + (x+1)^2 = 313
$$

Step 2: Expand
$$
x^2 + x^2 + 2x + 1 = 313 \\
2x^2 + 2x + 1 = 313
$$

Step 3: Simplify
$$
2x^2 + 2x - 312 = 0 \\
x^2 + x - 156 = 0
$$

Step 4: Factor
Find two numbers that multiply to $-156$ and add to $1$: $13$ and $-12$

$$
(x + 13)(x - 12) = 0
$$

So $ x = -13 $ or $ x = 12 $

Natural numbers → $ x = 12 $

Then the numbers are $ 12 $ and $ 13 $

Check: $ 12^2 + 13^2 = 144 + 169 = 313 $ ✔

✔ Answer: The numbers are 12 and 13.

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4) Two faucets can fill a tank in 1 hour and 20 mins. The time taken by faucet A alone to fill the tank is 2 hours more than faucet B. How long does it take faucet A alone to fill the tank?

Step 1: Convert time to minutes
1 hour 20 mins = 80 minutes

Let:
- Time taken by faucet B alone = $ x $ minutes
- Then time taken by faucet A alone = $ x + 120 $ minutes (since 2 hours = 120 minutes)

Step 2: Work rates
- Faucet A fills at rate: $ \frac{1}{x+120} $ tanks per minute
- Faucet B fills at rate: $ \frac{1}{x} $
- Together: $ \frac{1}{x+120} + \frac{1}{x} = \frac{1}{80} $

Step 3: Equation
$$
\frac{1}{x+120} + \frac{1}{x} = \frac{1}{80}
$$

Step 4: Multiply both sides by $ 80x(x+120) $ to eliminate denominators
$$
80x + 80(x+120) = x(x+120)
$$

Simplify:
$$
80x + 80x + 9600 = x^2 + 120x \\
160x + 9600 = x^2 + 120x
$$

Bring all terms to one side:
$$
0 = x^2 + 120x - 160x - 9600 \\
x^2 - 40x - 9600 = 0
$$

Step 5: Solve using quadratic formula
$$
x = \frac{40 \pm \sqrt{(-40)^2 + 4 \cdot 9600}}{2} = \frac{40 \pm \sqrt{1600 + 38400}}{2} = \frac{40 \pm \sqrt{40000}}{2}
$$
$$
\sqrt{40000} = 200 \Rightarrow x = \frac{40 \pm 200}{2}
$$

So:
- $ x = \frac{240}{2} = 120 $
- $ x = \frac{-160}{2} = -80 $ → discard

So $ x = 120 $ minutes → faucet B takes 120 min (2 hours)

Then faucet A takes $ 120 + 120 = 240 $ minutes = 4 hours

✔ Answer: Faucet A alone takes 4 hours to fill the tank.

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5) 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 three times the area of the square. Find the length of the side of the square.

Step 1: Let side of square = $ x $ cm

Area of square = $ x^2 $

New rectangle:
- Length = $ x + 10 $
- Width = $ x + 5 $
- Area = $ (x+10)(x+5) $

Given: Rectangle area = 3 × square area

So:
$$
(x+10)(x+5) = 3x^2
$$

Step 2: Expand left side
$$
x^2 + 15x + 50 = 3x^2
$$

Step 3: Bring all terms to one side
$$
0 = 3x^2 - x^2 - 15x - 50 = 2x^2 - 15x - 50
$$

Step 4: Solve quadratic
$$
2x^2 - 15x - 50 = 0
$$

Use quadratic formula:
$$
x = \frac{15 \pm \sqrt{(-15)^2 - 4(2)(-50)}}{2(2)} = \frac{15 \pm \sqrt{225 + 400}}{4} = \frac{15 \pm \sqrt{625}}{4}
$$
$$
\sqrt{625} = 25 \Rightarrow x = \frac{15 \pm 25}{4}
$$

So:
- $ x = \frac{40}{4} = 10 $
- $ x = \frac{-10}{4} = -2.5 $ → discard

✔ Answer: The side of the square is 10 cm.

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✔ Final Answers:

1) 8 seconds
2) 14 ft and 90 ft
3) 12 and 13
4) 4 hours
5) 10 cm

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