Question

As Isla prepares to set off fireworks, she wants to ensure the spectators are a safe distance away. When shot in the air, one firework will follow the pathh(x)=−x2+15x, where h(x) is the height of the firework, and x is the horizontal distance it travels in feet. To determine how far the firework will travel before it reaches the ground, Isla sets the height equal to zero, and solves the following quadratic equation: 0=−x2+15x. How far will the firework travel before it reaches the ground?(1 point)
Responses

15 feet
15 feet

−15 feet
negative 15 feet

30 feet
30 feet

7.5 feet

Answer 1

Hyung-eun is solving the quadratic equation x2−4x−12=0 by grouping. Her work is shown below. At which step does Hyung-eun first make an error?

x2−4x−12=0

Step1: x2+(−6x)+2x−12=0

Step 2: (x2+(−6x))+(2x−12)=0

Step 3: x(x+(−6))+2(x−6)=0

x(x−6)+2(x−6)=0

Step 4: (x+6)(x−2)=0

Step 5: x+6=0 or x−2=0

x=−6 or x=2

(1 point)
Responses

Step 4
Step 4

Step 3
Step 3

Step 2
Step 2

Step 1
Step 1
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Answer 2

Which of the following quadratic equations can be solved by grouping?(1 point)
Responses

x2+8x−22=0
x squared plus 8 x minus 22 equals 0

x2+10x+21=0
x squared plus 10 x plus 21 equals 0

x2−12x+18=0
x squared minus 12 x plus 18 equals 0

x2−4x−8=0

Answer 3

To find out how far the firework will travel before it reaches the ground, we need to solve the equation \(0 = -x^2 + 15x\).

First, rearranging the equation gives us:

\[
-x^2 + 15x = 0
\]

We can factor out \(x\):

\[
x(-x + 15) = 0
\]

This gives us two solutions:

1. Setting \(x = 0\), which represents the starting point when the firework is launched.
2. Setting \(-x + 15 = 0\), which simplifies to:

\[
-x = -15 \implies x = 15
\]

So, the firework will reach the ground when \(x = 15\) feet.

Therefore, the firework will travel **15 feet** before it reaches the ground. The correct response is:

**15 feet**

Answer 4

Let's analyze Hyung-eun's work step by step to identify where the mistake occurs.

1. **Original equation:**
\( x^2 - 4x - 12 = 0 \)

2. **Step 1:**
\( x^2 + (-6x) + 2x - 12 = 0 \)
This step is correct as it correctly rewrites \(-4x\) as \(-6x + 2x\).

3. **Step 2:**
\( (x^2 + (-6x)) + (2x - 12) = 0 \)
This step is also correct as it groups the terms properly.

4. **Step 3:**
\( x(x + (-6)) + 2(x - 6) = 0 \)
Here's where the first mistake appears. The first part of the factorization should be \( x(x - 6) \), not \( x(x + (-6)) \). Therefore, it should say \( x(x - 6) + 2(x - 6) = 0 \).

5. **Step 4:**
From the incorrect grouping in Step 3, Hyung-eun incorrectly writes \( (x + 6)(x - 2) = 0 \) instead of the correct \( (x - 6)(x + 2) = 0 \).

Thus, the error first occurs in **Step 3**, where she incorrectly factors the \( x^2 - 6x \) term.

The correct response is:

**Step 3**

Answer 5

To determine which quadratic equation can be solved by grouping, we should look for an equation where we can express the middle term as a sum of two terms that multiply to the product of the coefficient of \(x^2\) (which is 1 here) and the constant term.

In general, for a quadratic of the form \(ax^2 + bx + c\), we want to find two numbers that multiply to \(ac\) and add to \(b\).

Let's analyze each option:

1. **\(x^2 + 8x - 22 = 0\)**:
- Here, \(a = 1\), \(b = 8\), and \(c = -22\).
- The product \(ac = 1 \times -22 = -22\).
- We need two numbers that multiply to \(-22\) and add to \(8\). This does not work because there are no such pairs.

2. **\(x^2 + 10x + 21 = 0\)**:
- Here, \(a = 1\), \(b = 10\), and \(c = 21\).
- The product \(ac = 1 \times 21 = 21\).
- We need two numbers that multiply to \(21\) and add to \(10\). The numbers \(3\) and \(7\) work: \(3 \cdot 7 = 21\) and \(3 + 7 = 10\).
- This equation can be solved by grouping:
\[
x^2 + 3x + 7x + 21 = 0 \implies (x^2 + 3x) + (7x + 21) = 0.
\]

3. **\(x^2 - 12x + 18 = 0\)**:
- Here, \(a = 1\), \(b = -12\), and \(c = 18\).
- The product \(ac = 1 \times 18 = 18\).
- We need two numbers that multiply to \(18\) and add to \(-12\). The numbers do not work for the criteria together since they are both negative: \(-6\) and \(-3\) do not add to \(-12\).

4. **\(x^2 - 4x - 8 = 0\)**:
- Here, \(a = 1\), \(b = -4\), and \(c = -8\).
- The product \(ac = 1 \times -8 = -8\).
- We need two numbers that multiply to \(-8\) and add to \(-4\). The numbers \(2\) and \(-4\) do not work to add to \(-4\) but instead add to \(-2\).

Given all the evaluations:

The equation that can be solved by grouping is:

**\(x^2 + 10x + 21 = 0\)**