问题


There is a single choice question about high school mathematics. Answer the question by replying A, B, C or D.
Question: The quadratic $2x^2+5x+12=19-7x$ has two solutions. What is the positive difference between these solutions?
A. 2
B. \sqrt{2}
C. 5\sqrt{2}
D. 5
Answer:

答案

模型 模型答案 正确答案
金声玉亮2.0

B. \sqrt{2}
C
deepseekr1

To solve the quadratic equation \(2x^2 + 5x + 12 = 19 - 7x\), first rearrange all terms to one side:

\[
2x^2 + 5x + 12 - 19 + 7x = 0 \implies 2x^2 + 12x - 7 = 0.
\]

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 12\), and \(c = -7\):

1. Calculate the discriminant:
\[
\Delta = 12^2 - 4(2)(-7) = 144 + 56 = 200.
\]
\(\sqrt{200} = 10\sqrt{2}\).

2. Substitute into the formula:
\[
x = \frac{-12 \pm 10\sqrt{2}}{4} = \frac{-6 \pm 5\sqrt{2}}{2}.
\]

The solutions are \(\frac{-6 + 5\sqrt{2}}{2}\) and \(\frac{-6 - 5\sqrt{2}}{2}\). The positive difference between the roots is:

\[
\left|\frac{-6 + 5\sqrt{2}}{2} - \frac{-6 - 5\sqrt{2}}{2}\right| = \frac{10\sqrt{2}}{2} = 5\sqrt{2}.
\]

**Answer:** C. \(5\sqrt{2}\)
C
gpto1 C C
GPT-4o To find the positive difference between the solutions of the quadratic equation \(2x^2 + 5x + 12 = 19 - 7x\), we start by rewriting the equation in standard form.

First, simplify the equation:
\[2x^2 + 5x + 12 = 19 - 7x\]

Move all terms to one side to set the equation to zero:
\[2x^2 + 5x + 12 - 19 + 7x = 0\]

Combine like terms:
\[2x^2 + 12x - 7 = 0\]

Now, we have the quadratic equation in standard form:
\[2x^2 + 12x - 7 = 0\]

To find the positive difference between the solutions, we use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For the given quadratic equation \(2x^2 + 12x - 7 = 0\), the coefficients are:
\(a = 2\), \(b = 12\), and \(c = -7\).

First, find the discriminant:
\[\Delta = b^2 - 4ac\]
\[\Delta = 12^2 - 4(2)(-7)\]
\[\Delta = 144 + 56\]
\[\Delta = 200\]

Next, use the quadratic formula to find the roots:
\[x = \frac{-12 \pm \sqrt{200}}{2 \cdot 2}\]
\[x = \frac{-12 \pm \sqrt{200}}{4}\]
\[x = \frac{-12 \pm 10\sqrt{2}}{4}\]
\[x = \frac{-12}{4} \pm \frac{10\sqrt{2}}{4}\]
\[x = -3 \pm \frac{5\sqrt{2}}{2}\]

The solutions are:
\[x_1 = -3 + \frac{5\sqrt{2}}{2}\]
\[x_2 = -3 - \frac{5\sqrt{2}}{2}\]

To find the positive difference between these two solutions:
\[|x_1 - x_2| = \left| \left(-3 + \frac{5\sqrt{2}}{2}\right) - \left(-3 - \frac{5\sqrt{2}}{2}\right) \right|\]
\[|x_1 - x_2| = \left| -3 + \frac{5\sqrt{2}}{2} + 3 + \frac{5\sqrt{2}}{2} \right|\]
\[|x_1 - x_2| = \left| \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} \right|\]
\[|x_1 - x_2| = \left| \frac{10\sqrt{2}}{2} \right|\]
\[|x_1 - x_2| = 5\sqrt{2}\]

Thus, the positive difference between the solutions is:
\[C. 5\sqrt{2}\]
C
llama3.1-70B-Instruct

C
C
llama3.1-405B-Instruct

C
The best answer is C
C
DeepSeek-Coder-V2 To solve the quadratic equation \(2x^2 + 5x + 12 = 19 - 7x\), we first move all terms to one side to set the equation to zero:

\[
2x^2 + 5x + 12 - 19 + 7x = 0
\]

Simplify the equation:

\[
2x^2 + 12x - 7 = 0
\]

Next, we use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 12\), and \(c = -7\).

Calculate the discriminant \(\Delta = b^2 - 4ac\):

\[
\Delta = 12^2 - 4 \cdot 2 \cdot (-7) = 144 + 56 = 200
\]

Now, apply the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{200}}{4} = \frac{-12 \pm 10\sqrt{2}}{4} = \frac{-12}{4} \pm \frac{10\sqrt{2}}{4} = -3 \pm \frac{5\sqrt{2}}{2}
\]

The solutions are:

\[
x_1 = -3 + \frac{5\sqrt{2}}{2} \quad \text{and} \quad x_2 = -3 - \frac{5\sqrt{2}}{2}
\]

To find the positive difference between these solutions:

\[
\left| \left(-3 + \frac{5\sqrt{2}}{2}\right) - \left(-3 - \frac{5\sqrt{2}}{2}\right) \right| = \left| \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} \right| = \left| \frac{10\sqrt{2}}{2} \right| = 5\sqrt{2}
\]

Thus, the positive difference between the solutions is:

\[
\boxed{C}
\]
C
Qwen2-72B-Instruct First, we simplify the quadratic equation by moving all terms to one side:
\[2x^2 + 5x + 12 = 19 - 7x\]
\[2x^2 + 12x - 7 = 0\]
We can use the quadratic formula to find the solutions for x:
\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
Here, \(a = 2\), \(b = 12\), and \(c = -7\). Plugging these values into the formula gives:
\[x = \frac{-12 \pm \sqrt{12^2 - 4*2*(-7)}}{2*2}\]
\[x = \frac{-12 \pm \sqrt{144 + 56}}{4}\]
\[x = \frac{-12 \pm \sqrt{200}}{4}\]
\[x = \frac{-12 \pm 10\sqrt{2}}{4}\]
\[x = \frac{-12}{4} \pm \frac{10\sqrt{2}}{4}\]
\[x = -3 \pm \frac{5\sqrt{2}}{2}\]
So, the two solutions are:
\[x_1 = -3 + \frac{5\sqrt{2}}{2}\]
\[x_2 = -3 - \frac{5\sqrt{2}}{2}\]
The positive difference between these solutions is:
\[x_1 - x_2 = \left(-3 + \frac{5\sqrt{2}}{2}\right) - \left(-3 - \frac{5\sqrt{2}}{2}\right)\]
\[= \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}\]
\[= 5\sqrt{2}\]
Therefore, the answer is:
C. \(5\sqrt{2}\)
C
baichuan4 C C
doubao First, we need to transform the equation into the standard form of a quadratic equation: $2x^2 + 5x + 7x + 12 - 19 = 0$, which simplifies to $2x^2 + 12x - 7 = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 12$, and $c = -7$.

The discriminant is $12^2 - 4 \times 2 \times (-7) = 144 + 56 = 200$.

The two solutions are $x_1 = \frac{-12 + \sqrt{200}}{4} = \frac{-12 + 10\sqrt{2}}{4} = \frac{-3 + 5\sqrt{2}}{2}$ and $x_2 = \frac{-12 - \sqrt{200}}{4} = \frac{-12 - 10\sqrt{2}}{4} = \frac{-3 - 5\sqrt{2}}{2}$.

The positive difference between the solutions is $(\frac{-3 + 5\sqrt{2}}{2}) - (\frac{-3 - 5\sqrt{2}}{2}) = 5\sqrt{2}$.

So the answer is C.
C
minimax First, let's solve the quadratic equation by rearranging it into standard form:

\[2x^2 + 5x + 12 = 19 - 7x\]

Adding \(7x\) to both sides and subtracting \(19\) from both sides gives:

\[2x^2 + 12x - 7 = 0\]

Now we can use the quadratic formula to find the solutions for \(x\), where \(a = 2\), \(b = 12\), and \(c = -7\):

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Plugging in the values, we get:

\[x = \frac{-12 \pm \sqrt{12^2 - 4(2)(-7)}}{2(2)}\]
\[x = \frac{-12 \pm \sqrt{144 + 56}}{4}\]
\[x = \frac{-12 \pm \sqrt{200}}{4}\]
\[x = \frac{-12 \pm 10\sqrt{2}}{4}\]

Now we can simplify the solutions:

\[x_1 = \frac{-12 + 10\sqrt{2}}{4} = -3 + \frac{5\sqrt{2}}{2}\]
\[x_2 = \frac{-12 - 10\sqrt{2}}{4} = -3 - \frac{5\sqrt{2}}{2}\]

The positive difference between these two solutions is:

\[x_1 - x_2 = \left(-3 + \frac{5\sqrt{2}}{2}\right) - \left(-3 - \frac{5\sqrt{2}}{2}\right)\]
\[x_1 - x_2 = -3 + \frac{5\sqrt{2}}{2} + 3 + \frac{5\sqrt{2}}{2}\]
\[x_1 - x_2 = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}\]
\[x_1 - x_2 = 5\sqrt{2}\]

So the positive difference between the two solutions is \(5\sqrt{2}\), which corresponds to answer:

C. \(5\sqrt{2}\)
C
kimi First, let's simplify the equation by moving all terms to one side:

\[2x^2 + 5x + 12 = 19 - 7x\]

\[2x^2 + 5x + 7x + 12 - 19 = 0\]

\[2x^2 + 12x - 7 = 0\]

Now, we can divide the entire equation by 2 to simplify it further:

\[x^2 + 6x - \frac{7}{2} = 0\]

Next, we need to find the solutions to this quadratic equation. We can use the quadratic formula, which is given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For our equation, \(a = 1\), \(b = 6\), and \(c = -\frac{7}{2}\). Plugging these values into the quadratic formula gives us:

\[x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot \left(-\frac{7}{2}\right)}}{2 \cdot 1}\]

\[x = \frac{-6 \pm \sqrt{36 + 14}}{2}\]

\[x = \frac{-6 \pm \sqrt{50}}{2}\]

\[x = \frac{-6 \pm 5\sqrt{2}}{2}\]

\[x = -3 \pm \frac{5\sqrt{2}}{2}\]

The two solutions are \(x_1 = -3 + \frac{5\sqrt{2}}{2}\) and \(x_2 = -3 - \frac{5\sqrt{2}}{2}\).

The positive difference between these solutions is:

\[x_1 - x_2 = \left(-3 + \frac{5\sqrt{2}}{2}\right) - \left(-3 - \frac{5\sqrt{2}}{2}\right)\]

\[x_1 - x_2 = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}\]

\[x_1 - x_2 = 5\sqrt{2}\]

So the answer is C. \(5\sqrt{2}\).
C
wenxinyiyan C C
zhipu First, we need to bring the equation to standard quadratic form. Starting with the given equation:

\[ 2x^2 + 5x + 12 = 19 - 7x \]

Add \( 7x \) to both sides and subtract \( 19 \) from both sides to get:

\[ 2x^2 + 12x - 7 = 0 \]

Now, we solve the quadratic equation \( 2x^2 + 12x - 7 = 0 \) using the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 2 \), \( b = 12 \), and \( c = -7 \). Plugging in these values:

\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} \]

Calculate the discriminant:

\[ 12^2 - 4 \cdot 2 \cdot (-7) = 144 + 56 = 200 \]

So, the quadratic formula becomes:

\[ x = \frac{-12 \pm \sqrt{200}}{4} \]

Simplify \( \sqrt{200} \) to \( 10\sqrt{2} \):

\[ x = \frac{-12 \pm 10\sqrt{2}}{4} \]

Divide each term by 4:

\[ x = \frac{-12}{4} \pm \frac{10\sqrt{2}}{4} \]
\[ x = -3 \pm \frac{5\sqrt{2}}{2} \]

The two solutions are:

\[ x_1 = -3 + \frac{5\sqrt{2}}{2} \]
\[ x_2 = -3 - \frac{5\sqrt{2}}{2} \]

To find the positive difference between these solutions:

\[ \left| x_1 - x_2 \right| = \left| \left( -3 + \frac{5\sqrt{2}}{2} \right) - \left( -3 - \frac{5\sqrt{2}}{2} \right) \right| \]
\[ = \left| \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2} \right| \]
\[ = \left| 5\sqrt{2} \right| \]
\[ = 5\sqrt{2} \]

Therefore, the positive difference between the solutions is:

\[ \boxed{5\sqrt{2}} \]

So the correct answer is:

C. \( 5\sqrt{2} \)
C
gpto1 C C
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