Quadratic Equation worksheet with five word problems.

Here are the step-by-step solutions for each problem on the worksheet.

1) The Ball Problem

Problem: Abraham throws a ball from 40 m above the ground. Height is $h(t) = 40t - 5t^2$. How long until it hits the ground?

Step-by-step Solution:
1. When the ball hits the ground, the height $h(t)$ is equal to 0.
2. Set the equation to 0:
$$0 = 40t - 5t^2$$
3. Factor out the common term, which is $5t$:
$$0 = 5t(8 - t)$$
4. For this equation to be true, either part must be zero:
* $5t = 0 \rightarrow t = 0$ (This is when the ball was thrown)
* $8 - t = 0 \rightarrow t = 8$ (This is when it lands)

Answer: It takes 8 seconds.

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2) The Pool Dimensions

Problem: Area is $1260 \text{ ft}^2$. One side is 48 ft more than three times the other side. Find the dimensions.

Step-by-step Solution:
1. Let the shorter side be $x$.
2. The longer side is "48 more than 3 times $x$", so it is $(3x + 48)$.
3. Area = length $\times$ width. So:
$$x(3x + 48) = 1260$$
4. Multiply it out:
$$3x^2 + 48x = 1260$$
5. To make it easier, divide everything by 3:
$$x^2 + 16x = 420$$
6. Move 420 to the left to set it to zero:
$$x^2 + 16x - 420 = 0$$
7. We need two numbers that multiply to $-420$ and add to $+16$.
* Let's try factors of 420: $20 \times 21 = 420$. That's close.
* Let's try $30 \times 14 = 420$.
* Since we need $+16$, we use $+30$ and $-14$ ($30 - 14 = 16$).
8. So, $(x + 30)(x - 14) = 0$.
9. $x$ can be $-30$ or $14$. Since length can't be negative, $x = 14$.
10. Calculate the other side: $3(14) + 48 = 42 + 48 = 90$.

Answer: The dimensions are 14 ft and 90 ft.

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3) Consecutive Natural Numbers

Problem: Sum of squares of two consecutive natural numbers is 313. Find the numbers.

Step-by-step Solution:
1. Let the first number be $n$. The next consecutive number is $n + 1$.
2. Square them and add them:
$$n^2 + (n + 1)^2 = 313$$
3. Expand $(n + 1)^2$:
$$n^2 + (n^2 + 2n + 1) = 313$$
4. Combine like terms:
$$2n^2 + 2n + 1 = 313$$
5. Subtract 313 from both sides:
$$2n^2 + 2n - 312 = 0$$
6. Divide by 2 to simplify:
$$n^2 + n - 156 = 0$$
7. Find two numbers that multiply to $-156$ and add to $1$.
* $\sqrt{156}$ is roughly 12.5. Let's check numbers near 12.
* $12 \times 13 = 156$.
* We need positive 1, so use $+13$ and $-12$.
8. $(n + 13)(n - 12) = 0$.
9. $n = 12$ (we ignore -13 because natural numbers are positive).
10. The numbers are 12 and 13.
* Check: $12^2 + 13^2 = 144 + 169 = 313$. Correct.

Answer: The numbers are 12 and 13.

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4) The Faucets

Problem: Together they take 1 hr 20 mins. Faucet A takes 2 hours more than Faucet B alone. How long for Faucet A?

Step-by-step Solution:
1. Convert time to hours: 1 hour 20 mins = $1 + \frac{20}{60} = 1 \frac{1}{3}$ hours = $\frac{4}{3}$ hours.
2. Let $t$ be the time (in hours) it takes Faucet B to fill the tank alone.
3. Then Faucet A takes $t + 2$ hours.
4. Rate of work formula: $\frac{1}{A} + \frac{1}{B} = \frac{1}{\text{Total}}$
$$\frac{1}{t+2} + \frac{1}{t} = \frac{1}{4/3}$$
$$\frac{1}{t+2} + \frac{1}{t} = \frac{3}{4}$$
5. Multiply the whole equation by $4t(t+2)$ to clear denominators:
$$4(t) + 4(t+2) = 3t(t+2)$$
$$4t + 4t + 8 = 3t^2 + 6t$$
$$8t + 8 = 3t^2 + 6t$$
6. Rearrange into standard quadratic form ($ax^2+bx+c=0$):
$$0 = 3t^2 - 2t - 8$$
7. Factor the quadratic. We need numbers multiplying to $3 \times -8 = -24$ and adding to $-2$. Those are $-6$ and $4$.
$$3t^2 - 6t + 4t - 8 = 0$$
$$3t(t - 2) + 4(t - 2) = 0$$
$$(3t + 4)(t - 2) = 0$$
8. Solutions: $t = 2$ or $t = -\frac{4}{3}$. Time cannot be negative, so $t = 2$.
9. This means Faucet B takes 2 hours.
10. Faucet A takes $t + 2$, so $2 + 2 = 4$ hours.

Answer: It takes Faucet A 4 hours.

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5) The Square and Rectangle

Problem: Increase one side of a square by 10 cm, other by 5 cm. New area is 3 times the original square's area. Find the side of the square.

Step-by-step Solution:
1. Let the side of the square be $s$.
2. Original Area = $s^2$.
3. New dimensions are $(s + 10)$ and $(s + 5)$.
4. New Area = $(s + 10)(s + 5)$.
5. The problem says New Area = $3 \times$ Original Area:
$$(s + 10)(s + 5) = 3s^2$$
6. Expand the left side (FOIL method):
$$s^2 + 5s + 10s + 50 = 3s^2$$
$$s^2 + 15s + 50 = 3s^2$$
7. Move everything to the right side to keep $s^2$ positive:
$$0 = 2s^2 - 15s - 50$$
8. Factor $2s^2 - 15s - 50$.
* Multiply $a \cdot c = 2 \cdot -50 = -100$.
* Find factors of -100 that add to -15. They are -20 and +5.
* Rewrite middle term: $2s^2 - 20s + 5s - 50 = 0$.
* Group: $2s(s - 10) + 5(s - 10) = 0$.
* $(2s + 5)(s - 10) = 0$.
9. Solutions: $s = 10$ or $s = -2.5$. Length must be positive.
10. So, $s = 10$.

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

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