问题


There is a single choice question about high school mathematics. Answer the question by replying A, B, C or D.
Question: Rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, where $c$, $p$, and $q$ are constants. What is $\frac{q}{p}$?
A. 3
B. -34
C. -38
D. 11
Answer:

答案

模型 模型答案 正确答案
金声玉亮2.0

To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, we need to complete the square.

First, we factor out the coefficient of $j^2$, which is 6:

$$6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12$$

Next, we add and subtract $(\frac{2}{3\cdot 2})^2 = \frac{1}{9}$ inside the parentheses:

$$6(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}) + 12$$

Now, we can rewrite the expression as:

$$6(j^2 - \frac{2}{3}j + \frac{1}{9}) - \frac{6}{9} + 12$$

Simplifying, we get:

$$6(j - \frac{1}{3})^2 - \frac{2}{3} + 12$$

Combining the constant terms, we get:

$$6(j - \frac{1}{3})^2 + \frac{34}{3}$$

Comparing this with the desired form $c(j + p)^2 + q$, we see that:

$$c = 6, p = -\frac{1}{3}, q = \frac{34}{3}$$

Therefore, $\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34$.

The final answer is B.
B
deepseekr1

To rewrite the quadratic expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), we complete the square as follows:

1. **Factor out the coefficient of \(j^2\) from the first two terms**:
\[
6j^2 - 4j + 12 = 6\left(j^2 - \frac{2}{3}j\right) + 12.
\]

2. **Complete the square inside the parentheses**:
- The coefficient of \(j\) is \(-\frac{2}{3}\). Half of this coefficient is \(-\frac{1}{3}\), and squaring it gives \(\frac{1}{9}\).
- Add and subtract \(\frac{1}{9}\) inside the parentheses:
\[
6\left[\left(j^2 - \frac{2}{3}j + \frac{1}{9}\right) - \frac{1}{9}\right] + 12.
\]

3. **Simplify the expression**:
- The squared term becomes \(\left(j - \frac{1}{3}\right)^2\).
- Distribute the 6 and combine constants:
\[
6\left(j - \frac{1}{3}\right)^2 - 6 \cdot \frac{1}{9} + 12 = 6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12.
\]
- Combine \(- \frac{2}{3} + 12\):
\[
12 - \frac{2}{3} = \frac{36}{3} - \frac{2}{3} = \frac{34}{3}.
\]
- Final form: \(6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}\).

4. **Identify \(c\), \(p\), and \(q\)**:
- \(c = 6\), \(p = -\frac{1}{3}\) (since \(j + p = j - \frac{1}{3}\)), and \(q = \frac{34}{3}\).

5. **Calculate \(\frac{q}{p}\)**:
\[
\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34.
\]

**Answer: B**
B
gpto1 To solve the problem, we need to rewrite the quadratic expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\) and then calculate \(\frac{q}{p}\).

**Step 1: Factor out the coefficient of \(j^2\).**

Start by factoring out the 6:

\[6j^2 - 4j + 12 = 6\left(j^2 - \frac{2}{3}j\right) + 12\]

**Step 2: Complete the square inside the parentheses.**

To complete the square for \(j^2 - \frac{2}{3}j\):

- Take half of \(-\frac{2}{3}\): \(-\frac{1}{3}\)
- Square it: \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\)

Add and subtract \(\frac{1}{9}\) inside the parentheses:

\[6\left(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}\right) + 12\]

Simplify:

\[6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12\]

**Step 3: Distribute and simplify constants.**

Distribute 6:

\[6\left(j - \frac{1}{3}\right)^2 - 6\left(\frac{1}{9}\right) + 12\]

Simplify:

\[6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12\]

Combine constants:

\[-\frac{2}{3} + 12 = \frac{34}{3}\]

The expression becomes:

\[6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}\]

**Step 4: Identify \(c\), \(p\), and \(q\).**

From the expression:

- \(c = 6\)
- \(p = -\frac{1}{3}\)
- \(q = \frac{34}{3}\)

**Step 5: Calculate \(\frac{q}{p}\).**

\[\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = \frac{34}{3} \times \left(-\frac{3}{1}\right) = -34\]

**Conclusion:**

\(\frac{q}{p} = -34\), which corresponds to option **B**.

**Answer: B. -34**
B
GPT-4o To rewrite the expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), we will complete the square.

1. Start with the given expression:
\[
6j^2 - 4j + 12
\]

2. Factor out the coefficient of \(j^2\) from the first two terms:
\[
6(j^2 - \frac{2}{3}j) + 12
\]

3. Complete the square inside the parentheses. To do this, take half of the coefficient of \(j\) (which is \(-\frac{2}{3}\)), square it, and add and subtract it inside the parentheses:
\[
j^2 - \frac{2}{3}j \quad \text{half of } -\frac{2}{3} \text{ is } -\frac{1}{3}, \text{ and } \left(-\frac{1}{3}\right)^2 = \frac{1}{9}
\]
\[
6\left(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}\right) + 12
\]
\[
6\left((j - \frac{1}{3})^2 - \frac{1}{9}\right) + 12
\]

4. Distribute the 6:
\[
6(j - \frac{1}{3})^2 - 6 \cdot \frac{1}{9} + 12
\]
\[
6(j - \frac{1}{3})^2 - \frac{6}{9} + 12
\]
\[
6(j - \frac{1}{3})^2 - \frac{2}{3} + 12
\]
\[
6(j - \frac{1}{3})^2 + \frac{34}{3}
\]

So the expression \(6j^2 - 4j + 12\) can be rewritten as:
\[
6(j - \frac{1}{3})^2 + \frac{34}{3}
\]

In this form, \(c = 6\), \(p = -\frac{1}{3}\), and \(q = \frac{34}{3}\).

Now we need to find \(\frac{q}{p}\):
\[
\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34
\]

Therefore, the answer is:
\[
B. -34
\]
B
llama3.1-70B-Instruct

B
The best answer is B.
B
llama3.1-405B-Instruct

To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, we need to complete the square.

First, we factor out the coefficient of $j^2$, which is 6:

$$6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12$$

Next, we add and subtract the square of half the coefficient of $j$ inside the parentheses:

$$6(j^2 - \frac{2}{3}j + \frac{1}{9}) + 12 - \frac{6}{9}$$

Simplifying, we get:

$$6(j - \frac{1}{3})^2 + 12 - \frac{2}{3}$$

$$6(j - \frac{1}{3})^2 + \frac{34}{3}$$

Now we can identify the values of $c$, $p$, and $q$:

$$c = 6, p = -\frac{1}{3}, q = \frac{34}{3}$$

Finally, we can calculate $\frac{q}{p}$:

$$\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34$$

The final answer is B.
B
DeepSeek-Coder-V2 To rewrite the expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), we need to complete the square. Here are the steps:

1. **Factor out the coefficient of \(j^2\) from the first two terms:**
\[
6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12
\]

2. **Complete the square inside the parentheses:**
- Take the coefficient of \(j\), which is \(-\frac{2}{3}\), divide it by 2, and square it:
\[
\left(\frac{-\frac{2}{3}}{2}\right)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9}
\]
- Add and subtract this square inside the parentheses:
\[
6\left(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}\right) + 12 = 6\left((j - \frac{1}{3})^2 - \frac{1}{9}\right) + 12
\]

3. **Simplify the expression:**
\[
6\left((j - \frac{1}{3})^2 - \frac{1}{9}\right) + 12 = 6(j - \frac{1}{3})^2 - 6 \cdot \frac{1}{9} + 12 = 6(j - \frac{1}{3})^2 - \frac{2}{3} + 12
\]
\[
= 6(j - \frac{1}{3})^2 + \frac{34}{3}
\]

Now, the expression is in the form \(c(j + p)^2 + q\) where \(c = 6\), \(p = -\frac{1}{3}\), and \(q = \frac{34}{3}\).

4. **Calculate \(\frac{q}{p}\):**
\[
\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = \frac{34}{3} \cdot \frac{3}{-1} = -34
\]

Thus, the answer is:
\[
\boxed{B}
\]
B
Qwen2-72B-Instruct To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, let's complete the square.

The expression is $6j^2 - 4j + 12$. First, factor out the coefficient of $j^2$ (which is 6) from the first two terms:
\[6(j^2 - \frac{2}{3}j) + 12\]

To complete the square, add and subtract the square of half the coefficient of $j$ inside the parentheses, which is $\left(\frac{-\frac{2}{3}}{2}\right)^2 = \left(\frac{-2}{6}\right)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9}$:
\[6\left(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}\right) + 12\]

Simplify:
\[6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12\]

Distribute the 6:
\[6\left(j - \frac{1}{3}\right)^2 - 6\cdot\frac{1}{9} + 12\]

\[6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12\]

\[6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}\]

So, we have $c = 6$, $p = -\frac{1}{3}$, and $q = \frac{34}{3}$.

To find $\frac{q}{p}$, we have:
\[\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34\]

Therefore, the correct answer is B. -34.
B
baichuan4 To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, we first need to complete the square.

The original expression can be written as:
$$6(j^2 - \frac{2}{3}j) + 12$$

In order to complete the square, we add and subtract the square of half the coefficient of $j$, which is $\left(\frac{-1}{3}\right)^2 = \frac{1}{9}$:
$$6\left(j^2 - \frac{2}{3}j + \frac{1}{9}\right) + 12 - \frac{6}{9}$$

This can be simplified to:
$$6(j - \frac{1}{3})^2 + 12 - \frac{2}{3}$$
$$= 6(j - \frac{1}{3})^2 + \frac{36}{3} - \frac{2}{3}$$
$$= 6(j - \frac{1}{3})^2 + \frac{34}{3}$$

So, we have $c = 6$, $p = -\frac{1}{3}$, and $q = \frac{34}{3}$.

Now, to find $\frac{q}{p}$:
$$\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34$$

Therefore, the answer is B. -34.
B
doubao 6j^2 - 4j + 12 = 6(j^2 - 2/3j) + 12 = 6(j^2 - 2/3j + 1/9 - 1/9) + 12 = 6[(j - 1/3)^2 - 1/9] + 12 = 6(j - 1/3)^2 - 2/3 + 12 = 6(j - 1/3)^2 + 34/3. So p = -1/3, q = 34/3, and q / p = (34/3) / (-1/3) = -34. The answer is B. B
minimax To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, we need to complete the square.

First, we factor out the coefficient of $j^2$ from the first two terms:

$6(j^2 - \frac{2}{3}j) + 12$

Now, to complete the square, we need to add and subtract the square of half the coefficient of $j$ inside the parentheses:

The coefficient of $j$ inside the parentheses is $-\frac{2}{3}$, so half of this is $-\frac{1}{3}$, and squaring this gives us $\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$.

We add and subtract $6 \cdot \frac{1}{9}$ inside the parentheses:

$6\left(j^2 - \frac{2}{3}j + \frac{1}{9}\right) - 6 \cdot \frac{1}{9} + 12$

Now we can rewrite the expression inside the parentheses as a perfect square:

$6\left(j - \frac{1}{3}\right)^2 - \frac{6}{9} + 12$

Simplify the constants:

$6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12$

$6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}$

So, we have $c = 6$, $p = -\frac{1}{3}$, and $q = \frac{34}{3}$.

Now we calculate $\frac{q}{p}$:

$\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34$

The answer is B. -34.
B
kimi To rewrite the expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), we can complete the square.

First, factor out the coefficient of \(j^2\) from the first two terms:
\[6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12\]

Next, complete the square inside the parentheses:
\[j^2 - \frac{2}{3}j = \left(j - \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2\]

Now, substitute back into the original expression:
\[6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12\]

Distribute the 6:
\[6\left(j - \frac{1}{3}\right)^2 - \frac{6}{9} + 12\]

Simplify the constants:
\[6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12\]

Convert 12 to a fraction with a denominator of 3:
\[6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + \frac{36}{3}\]

Combine the constants:
\[6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}\]

So, we have \(c = 6\), \(p = -\frac{1}{3}\), and \(q = \frac{34}{3}\).

Now, we need to find \(\frac{q}{p}\):
\[\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = \frac{34}{3} \times \left(-\frac{3}{1}\right) = -34\]

The answer is B. -34.
B
wenxinyiyan B

Explanation:
To rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, we can use the method of completing the square.

First, we factor out the coefficient of $j^2$, which is 6:
$6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12$

Next, we find the value of $(\frac{1}{2} \cdot -\frac{2}{3})^2 = \frac{1}{9}$ to complete the square. We add and subtract this value inside the parentheses:
$6(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}) + 12$

Now we can write the expression as a perfect square:
$6((j - \frac{1}{3})^2 - \frac{1}{9}) + 12$

We distribute the 6 back into the parentheses:
$6(j - \frac{1}{3})^2 - \frac{2}{3} + 12$

Combine the constant terms:
$6(j - \frac{1}{3})^2 + 12 - \frac{2}{3}$
$6(j - \frac{1}{3})^2 + \frac{36}{3} - \frac{2}{3}$
$6(j - \frac{1}{3})^2 + \frac{34}{3}$

Now we have the expression in the desired form, with $c = 6$, $p = -\frac{1}{3}$, and $q = \frac{34}{3}$.

Finally, we find $\frac{q}{p}$:
$\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34$
B
zhipu To rewrite the expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), we need to complete the square.

First, factor out the coefficient of \(j^2\) from the first two terms:
\[ 6j^2 - 4j + 12 = 6(j^2 - \frac{2}{3}j) + 12 \]

Next, complete the square inside the parentheses:
\[ j^2 - \frac{2}{3}j \]
To complete the square, add and subtract \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\) inside the parentheses:
\[ j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9} = \left(j - \frac{1}{3}\right)^2 - \frac{1}{9} \]

Now substitute back into the original expression:
\[ 6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12 \]
Distribute the 6:
\[ 6\left(j - \frac{1}{3}\right)^2 - 6 \cdot \frac{1}{9} + 12 \]
\[ 6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12 \]

Combine the constants:
\[ 6\left(j - \frac{1}{3}\right)^2 + 12 - \frac{2}{3} \]
\[ 6\left(j - \frac{1}{3}\right)^2 + \frac{36}{3} - \frac{2}{3} \]
\[ 6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3} \]

Now, we have the expression in the form \(c(j + p)^2 + q\), where:
\[ c = 6, \quad p = -\frac{1}{3}, \quad q = \frac{34}{3} \]

We need to find \(\frac{q}{p}\):
\[ \frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = \frac{34}{3} \times \left(-3\right) = -34 \]

Thus, the answer is:
\[ \boxed{-34} \]

So the correct choice is:
B. -34
B
gpto1 To solve the problem, we need to rewrite the quadratic expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\) and then calculate \(\frac{q}{p}\).

**Step 1: Factor out the coefficient of \(j^2\).**

Start by factoring out the 6:

\[6j^2 - 4j + 12 = 6\left(j^2 - \frac{2}{3}j\right) + 12\]

**Step 2: Complete the square inside the parentheses.**

To complete the square for \(j^2 - \frac{2}{3}j\):

- Take half of \(-\frac{2}{3}\): \(-\frac{1}{3}\)
- Square it: \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\)

Add and subtract \(\frac{1}{9}\) inside the parentheses:

\[6\left(j^2 - \frac{2}{3}j + \frac{1}{9} - \frac{1}{9}\right) + 12\]

Simplify:

\[6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12\]

**Step 3: Distribute and simplify constants.**

Distribute 6:

\[6\left(j - \frac{1}{3}\right)^2 - 6\left(\frac{1}{9}\right) + 12\]

Simplify:

\[6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12\]

Combine constants:

\[-\frac{2}{3} + 12 = \frac{34}{3}\]

The expression becomes:

\[6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}\]

**Step 4: Identify \(c\), \(p\), and \(q\).**

From the expression:

- \(c = 6\)
- \(p = -\frac{1}{3}\)
- \(q = \frac{34}{3}\)

**Step 5: Calculate \(\frac{q}{p}\).**

\[\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = \frac{34}{3} \times \left(-\frac{3}{1}\right) = -34\]

**Conclusion:**

\(\frac{q}{p} = -34\), which corresponds to option **B**.

**Answer: B. -34**
B
deepseekv3 To rewrite the expression \(6j^2 - 4j + 12\) in the form \(c(j + p)^2 + q\), follow these steps:

1. **Factor out the coefficient of \(j^2\) from the first two terms:**
\[
6j^2 - 4j + 12 = 6\left(j^2 - \frac{2}{3}j\right) + 12
\]

2. **Complete the square inside the parentheses:**
\[
j^2 - \frac{2}{3}j
\]
To complete the square, take half of the coefficient of \(j\), square it, and add and subtract it inside the parentheses:
\[
j^2 - \frac{2}{3}j + \left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 = \left(j - \frac{1}{3}\right)^2 - \frac{1}{9}
\]

3. **Substitute back into the expression:**
\[
6\left(\left(j - \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 12 = 6\left(j - \frac{1}{3}\right)^2 - \frac{6}{9} + 12
\]
Simplify the constants:
\[
6\left(j - \frac{1}{3}\right)^2 - \frac{2}{3} + 12 = 6\left(j - \frac{1}{3}\right)^2 + \frac{34}{3}
\]

4. **Identify the constants:**
\[
c = 6, \quad p = -\frac{1}{3}, \quad q = \frac{34}{3}
\]

5. **Calculate \(\frac{q}{p}\):**
\[
\frac{q}{p} = \frac{\frac{34}{3}}{-\frac{1}{3}} = -34
\]

**Answer:** \(\boxed{B}\)
B