| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 12x^2 + 2x + 11 = 12 \left( x^2 + \frac{2}{12}x \right) + 11 \] Simplify the fraction inside the parentheses: \[ \frac{2}{12} = \frac{1}{6} \] So, the expression becomes: \[ 12 \left( x^2 + \frac{1}{6}x \right) + 11 \] 2. **Complete the square inside the parentheses:** To complete the square, we need to add and subtract the square of half the coefficient of \(x\). The coefficient of \(x\) is \(\frac{1}{6}\), so half of this is \(\frac{1}{12}\), and its square is: \[ \left( \frac{1}{12} \right)^2 = \frac{1}{144} \] Add and subtract \(\frac{1}{144}\) inside the parentheses: \[ 12 \left( x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144} \right) + 11 \] This can be rewritten as: \[ 12 \left( x^2 + \frac{1}{6}x + \frac{1}{144} \right) - 12 \cdot \frac{1}{144} + 11 \] 3. **Simplify the expression:** The term inside the parentheses is now a perfect square: \[ x^2 + \frac{1}{6}x + \frac{1}{144} = \left( x + \frac{1}{12} \right)^2 \] So, the expression becomes: \[ 12 \left( x + \frac{1}{12} \right)^2 - 12 \cdot \frac{1}{144} + 11 \] Simplify the constant term: \[ -12 \cdot \frac{1}{144} = -\frac{12}{144} = -\frac{1}{12} \] Therefore, the expression is: \[ 12 \left( x + \frac{1}{12} \right)^2 - \frac{1}{12} + 11 \] 4. **Combine the constants:** \[ -\frac{1}{12} + 11 = -\frac{1}{12} + \frac{132}{12} = \frac{131}{12} \] So, the completed square form of the quadratic expression \(12x^2 + 2x + 11\) is: \[ 12 \left( x + \frac{1}{12} \right)^2 + \frac{131}{12} \] Putting the final answer in a \(\boxed{}\): \[ \boxed{12 \left( x + \frac{1}{12} \right)^2 + \frac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| deepseek-r1 | To complete the square for the quadratic \(12x^2 + 2x + 11\), we start by factoring out the coefficient of \(x^2\) from the first two terms: \[ 12x^2 + 2x + 11 = 12\left(x^2 + \frac{1}{6}x\right) + 11 \] Next, we complete the square inside the parentheses. The coefficient of \(x\) is \(\frac{1}{6}\). We take half of this coefficient, which is \(\frac{1}{12}\), and square it to get \(\left(\frac{1}{12}\right)^2 = \frac{1}{144}\). We add and subtract this value inside the parentheses: \[ 12\left[\left(x^2 + \frac{1}{6}x + \frac{1}{144}\right) - \frac{1}{144}\right] + 11 \] This can be rewritten as: \[ 12\left[\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right] + 11 \] Distributing the 12 through the terms inside the brackets, we get: \[ 12\left(x + \frac{1}{12}\right)^2 - 12 \cdot \frac{1}{144} + 11 \] Simplifying the constants: \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] Converting 11 to twelfths to combine the constants: \[ 12\left(x + \frac{1}{12}\right)^2 + \left(\frac{132}{12} - \frac{1}{12}\right) = 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] Thus, the quadratic in completed square form is: \[ \boxed{12\left(x + \dfrac{1}{12}\right)^2 + \dfrac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| gpto1 | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we'll follow a systematic approach. **Step 1**: Factor out the coefficient of \(x^2\) from the first two terms. \[ 12x^2 + 2x = 12\left(x^2 + \frac{1}{6}x\right) \] **Step 2**: Complete the square inside the parentheses. To do this, take half of the coefficient of \(x\) inside the parentheses and square it. Half of \(\frac{1}{6}\) is \(\frac{1}{12}\). Square it to get \(\left(\frac{1}{12}\right)^2 = \frac{1}{144}\). **Step 3**: Add and subtract this squared term inside the parentheses (so we effectively add zero). \[ 12\left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + 11 \] **Step 4**: Write the perfect square trinomial and simplify. \[ 12\left[\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right] + 11 \] Distribute the 12: \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{12}{144} + 11 \] Simplify \(-\frac{12}{144}\) to \(-\frac{1}{12}\): \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] **Step 5**: Combine the constants. \[ -\frac{1}{12} + 11 = -\frac{1}{12} + \frac{132}{12} = \frac{131}{12} \] **Step 6**: Write the final expression. \[ 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] So, the completed square form of the quadratic \(12x^2 + 2x + 11\) is: \[ \boxed{12\left(x + \dfrac{1}{12}\right)^2 + \dfrac{131}{12}} \] \(\boxed{12\left(x+\dfrac{1}{12}\right)^{2}+\dfrac{131}{12}}\) |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| gpt-4o-2024-08-06 | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we first need to factor out the coefficient of \(x^2\) from the first two terms. The coefficient of \(x^2\) is 12, so we factor 12 out: \[ 12x^2 + 2x + 11 = 12(x^2 + \frac{1}{6}x) + 11 \] Next, we complete the square inside the parentheses. The expression inside is \(x^2 + \frac{1}{6}x\). To complete the square, we take half of the coefficient of \(x\), which is \(\frac{1}{6}\), and square it: \[ \left(\frac{1}{2} \times \frac{1}{6}\right)^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144} \] We add and subtract this square inside the parentheses: \[ x^2 + \frac{1}{6}x = \left(x^2 + \frac{1}{6}x + \frac{1}{144}\right) - \frac{1}{144} \] This can be rewritten as a perfect square: \[ x^2 + \frac{1}{6}x + \frac{1}{144} = \left(x + \frac{1}{12}\right)^2 \] Thus, the expression becomes: \[ 12\left(x^2 + \frac{1}{6}x\right) = 12\left(\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right) \] Distribute the 12: \[ 12\left(x + \frac{1}{12}\right)^2 - 12 \times \frac{1}{144} = 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} \] Now, add the constant term 11 back: \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] Combine the constants: \[ -\frac{1}{12} + 11 = -\frac{1}{12} + \frac{132}{12} = \frac{131}{12} \] Thus, the completed square form of the quadratic is: \[ 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] The final answer is: \[ \boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| DeepSeek-Coder-V2-Lite-Instruct | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 12x^2 + 2x + 11 = 12 \left(x^2 + \frac{2}{12}x\right) + 11 \] Simplify the fraction: \[ 12 \left(x^2 + \frac{1}{6}x\right) + 11 \] 2. **Complete the square inside the parentheses:** To complete the square, we need to add and subtract the square of half the coefficient of \(x\) inside the parentheses. The coefficient of \(x\) is \(\frac{1}{6}\), so half of this is \(\frac{1}{12}\), and its square is \(\left(\frac{1}{12}\right)^2 = \frac{1}{144}\). Add and subtract \(\frac{1}{144}\) inside the parentheses: \[ 12 \left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + 11 \] This simplifies to: \[ 12 \left(\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right) + 11 \] 3. **Distribute the 12 and simplify:** \[ 12 \left(x + \frac{1}{12}\right)^2 - 12 \cdot \frac{1}{144} + 11 \] Simplify the multiplication: \[ 12 \left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] 4. **Combine the constants:** \[ 12 \left(x + \frac{1}{12}\right)^2 + \left(11 - \frac{1}{12}\right) \] Convert 11 to a fraction with a common denominator: \[ 11 = \frac{132}{12} \] Subtract the fractions: \[ 11 - \frac{1}{12} = \frac{132}{12} - \frac{1}{12} = \frac{131}{12} \] 5. **Write the final expression:** \[ 12 \left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] Thus, the completed square form of the quadratic expression \(12x^2 + 2x + 11\) is: \[ \boxed{12 \left(x + \frac{1}{12}\right)^2 + \frac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $12x^2 + 2x + 11$. The coefficients are $a = 12$, $b = 2$, and $c = 11$. ## Step 2: Divide the entire equation by the coefficient of $x^2$ to make the coefficient of $x^2$ equal to 1 To simplify the process, divide the entire equation by $a = 12$ to get $x^2 + \frac{2}{12}x + \frac{11}{12}$. ## Step 3: Move the constant term to the other side of the equation Since we're completing the square, we'll focus on the terms involving $x$. The equation becomes $x^2 + \frac{2}{12}x = -\frac{11}{12}$. ## Step 4: Find the value to complete the square To complete the square, we take half of the coefficient of $x$, square it, and add it to both sides. The coefficient of $x$ is $\frac{2}{12}$, so half of it is $\frac{1}{12}$. Squaring this gives $\left(\frac{1}{12}\right)^2 = \frac{1}{144}$. ## Step 5: Add the value to complete the square to both sides of the equation Adding $\frac{1}{144}$ to both sides gives $x^2 + \frac{2}{12}x + \frac{1}{144} = -\frac{11}{12} + \frac{1}{144}$. ## Step 6: Simplify the right side of the equation Combining the fractions on the right side gives $-\frac{11}{12} + \frac{1}{144} = -\frac{11 \times 12}{12 \times 12} + \frac{1}{144} = -\frac{132}{144} + \frac{1}{144} = -\frac{131}{144}$. ## Step 7: Write the completed square The left side of the equation is now a perfect square: $\left(x + \frac{1}{12}\right)^2 = -\frac{131}{144}$. ## Step 8: Write the final form of the completed square Since the original request was to complete the square for the quadratic and the given quadratic was in the form $ax^2 + bx + c$, the completed square form should be presented in a way that matches the original format, adjusting for the process we've done. However, the traditional method of completing the square doesn't change the original expression's form into a perfect square trinomial directly but rather transforms it into a form that reveals the roots or the vertex of the parabola more clearly. Given the steps followed, the focus was on demonstrating the process rather than converting the original quadratic into a perfect square trinomial form directly. The final answer is: $\boxed{12\left(x + \frac{1}{12}\right)^2 -\frac{131}{12}}$ |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $12x^2 + 2x + 11$. The coefficients are $a = 12$, $b = 2$, and $c = 11$. ## Step 2: Determine the coefficient of the squared term The coefficient of the squared term is $a = 12$. To complete the square, we need to factor this coefficient out of the first two terms. ## Step 3: Factor out the coefficient of the squared term Factoring out $12$ from the first two terms gives $12(x^2 + \frac{1}{6}x) + 11$. ## Step 4: Calculate the value to complete the square To complete the square, we need to add and subtract $(\frac{b}{2})^2$ inside the parentheses. In this case, $\frac{b}{2} = \frac{1}{12}$, so $(\frac{b}{2})^2 = \frac{1}{144}$. ## Step 5: Add and subtract the calculated value Adding and subtracting $\frac{1}{144}$ inside the parentheses gives $12(x^2 + \frac{1}{6}x + \frac{1}{144}) + 11 - \frac{1}{12}$. ## Step 6: Simplify the expression Simplifying the expression gives $12(x + \frac{1}{12})^2 + \frac{131}{12}$. ## Step 7: Write the final answer in the required format The final answer is: $\boxed{12(x + \frac{1}{12})^2 + \frac{131}{12}}$ |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| Qwen2-72B-Instruct | To complete the square for the quadratic expression $12x^2 + 2x + 11$, we need to follow these steps: 1. Factor out the coefficient of $x^2$ (which is 12) from the first two terms. 2. Complete the square inside the parentheses by adding and subtracting the square of half the coefficient of $x$ divided by the coefficient of $x^2$. 3. Simplify the expression. Let's do this step by step: \[ 12x^2 + 2x + 11 = 12(x^2 + \frac{2}{12}x) + 11 \] Simplify the fraction $\frac{2}{12}$ to $\frac{1}{6}$: \[ = 12(x^2 + \frac{1}{6}x) + 11 \] Now, add and subtract the square of half the coefficient of $x$ inside the parentheses. The coefficient of $x$ is $\frac{1}{6}$, so half of that is $\frac{1}{12}$, and its square is $(\frac{1}{12})^2 = \frac{1}{144}$: \[ = 12\left(x^2 + \frac{1}{6}x + (\frac{1}{12})^2 - (\frac{1}{12})^2\right) + 11 \] Since we're adding and subtracting the same value, we can rewrite it as: \[ = 12\left(x^2 + \frac{1}{6}x + (\frac{1}{12})^2\right) - 12(\frac{1}{12})^2 + 11 \] Simplify the expression: \[ = 12\left(x + \frac{1}{12}\right)^2 - 12(\frac{1}{144}) + 11 \] Calculate $- 12(\frac{1}{144})$: \[ = 12\left(x + \frac{1}{12}\right)^2 - \frac{12}{144} + 11 \] Simplify $\frac{12}{144}$ to $\frac{1}{12}$: \[ = 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] Convert 11 to a fraction with a denominator of 12 to combine the fractions: \[ = 12\left(x + \frac{1}{12}\right)^2 + \frac{132}{12} - \frac{1}{12} \] Combine the fractions: \[ = 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] So, the completed square form of $12x^2 + 2x + 11$ is: \[ \boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| baichuan4 | To complete the square for the quadratic $12x^2 + 2x + 11$, we need to follow a few steps. The goal is to rewrite the quadratic in the form $(ax + b)^2 + c$, where $a$, $b$, and $c$ are constants. First, we factor out the coefficient of $x^2$, which is 12, to make the quadratic look like a perfect square trinomial: $$12x^2 + 2x + 11 = 12(x^2 + \frac{1}{6}x) + 11.$$ Next, we need to determine the constant that we need to add inside the parentheses to make $x^2 + \frac{1}{6}x$ a perfect square. This constant is half the coefficient of $x$ (from the linear term), squared. In this case, the coefficient of $x$ is $\frac{1}{6}$, so half of that is $\frac{1}{12}$. Squared, this gives us $\left(\frac{1}{12}\right)^2 = \frac{1}{144}$. We add this constant inside the parentheses, but because we are adding it inside the parentheses, we also need to multiply it by the coefficient we factored out initially (which was 12). So, we actually add $12 \cdot \frac{1}{144} = \frac{1}{12}$ outside the parentheses to balance the equation: $$12(x^2 + \frac{1}{6}x + \frac{1}{144}) + 11 + \frac{1}{12}.$$ Now, we can rewrite the expression inside the parentheses as a perfect square: $$12\left(x + \frac{1}{12}\right)^2 + 11 + \frac{1}{12}.$$ Finally, we simplify the constants outside the parentheses: $$12\left(x + \frac{1}{12}\right)^2 + \frac{133}{12}.$$ This is the completed square form of the original quadratic. The final answer is: $$\boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{133}{12}}.$$ |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| doubao | $12x^2+2x+11$ $=12(x^2+\frac{2}{12}x)+11$ $=12(x^2+\frac{1}{6}x)+11$ $=12(x^2+\frac{1}{6}x+\frac{1}{144}-\frac{1}{144})+11$ $=12((x+\frac{1}{12})^2-\frac{1}{144})+11$ $=12(x+\frac{1}{12})^2-12\times\frac{1}{144}+11$ $=12(x+\frac{1}{12})^2-\frac{1}{12}+11$ $=12(x+\frac{1}{12})^2+\frac{131}{12}$ $\boxed{12(x+\frac{1}{12})^2+\frac{131}{12}}$ |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| gpto1 | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we'll follow a systematic approach. **Step 1**: Factor out the coefficient of \(x^2\) from the first two terms. \[ 12x^2 + 2x = 12\left(x^2 + \frac{1}{6}x\right) \] **Step 2**: Complete the square inside the parentheses. To do this, take half of the coefficient of \(x\) inside the parentheses and square it. Half of \(\frac{1}{6}\) is \(\frac{1}{12}\). Square it to get \(\left(\frac{1}{12}\right)^2 = \frac{1}{144}\). **Step 3**: Add and subtract this squared term inside the parentheses (so we effectively add zero). \[ 12\left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + 11 \] **Step 4**: Write the perfect square trinomial and simplify. \[ 12\left[\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right] + 11 \] Distribute the 12: \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{12}{144} + 11 \] Simplify \(-\frac{12}{144}\) to \(-\frac{1}{12}\): \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] **Step 5**: Combine the constants. \[ -\frac{1}{12} + 11 = -\frac{1}{12} + \frac{132}{12} = \frac{131}{12} \] **Step 6**: Write the final expression. \[ 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] So, the completed square form of the quadratic \(12x^2 + 2x + 11\) is: \[ \boxed{12\left(x + \dfrac{1}{12}\right)^2 + \dfrac{131}{12}} \] \(\boxed{12\left(x+\dfrac{1}{12}\right)^{2}+\dfrac{131}{12}}\) |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| kimi | To complete the square for the quadratic expression \(12x^2 + 2x + 11\), we will follow these steps: 1. **Factor out the coefficient of \(x^2\) from the first two terms:** \[ 12x^2 + 2x + 11 = 12(x^2 + \frac{1}{6}x) + 11 \] 2. **Complete the square inside the parentheses:** - Take the coefficient of \(x\), which is \(\frac{1}{6}\). - Square it: \(\left(\frac{1}{6}\right)^2 = \frac{1}{36}\). - Add and subtract this square inside the parentheses: \[ x^2 + \frac{1}{6}x = x^2 + \frac{1}{6}x + \frac{1}{36} - \frac{1}{36} \] - Rewrite it as a perfect square trinomial: \[ x^2 + \frac{1}{6}x + \frac{1}{36} - \frac{1}{36} = \left(x + \frac{1}{12}\right)^2 - \frac{1}{36} \] 3. **Substitute back into the original expression:** \[ 12\left(x^2 + \frac{1}{6}x\right) + 11 = 12\left(\left(x + \frac{1}{12}\right)^2 - \frac{1}{36}\right) + 11 \] 4. **Distribute the 12:** \[ 12\left(x + \frac{1}{12}\right)^2 - 12 \cdot \frac{1}{36} + 11 \] Simplify \(12 \cdot \frac{1}{36}\): \[ 12 \cdot \frac{1}{36} = \frac{12}{36} = \frac{1}{3} \] So the expression becomes: \[ 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{3} + 11 \] 5. **Combine the constant terms:** \[ -\frac{1}{3} + 11 = 11 - \frac{1}{3} = \frac{33}{3} - \frac{1}{3} = \frac{32}{3} \] 6. **Write the final expression:** \[ 12\left(x + \frac{1}{12}\right)^2 + \frac{32}{3} \] Thus, the completed square form of the quadratic \(12x^2 + 2x + 11\) is: \[ \boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{32}{3}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| minimax | To complete the square for the quadratic equation $12x^2 + 2x + 11$, we need to express it in the form $(ax + b)^2 + c$. First, we factor out the coefficient of $x^2$ from the first two terms: $$12(x^2 + \frac{2}{12}x) + 11$$ Simplify the fraction: $$12(x^2 + \frac{1}{6}x) + 11$$ Now, to complete the square, we need to add and subtract the square of half the coefficient of $x$ inside the parentheses. The coefficient of $x$ inside the parentheses is $\frac{1}{6}$, so half of that is $\frac{1}{12}$, and squaring that gives us $\left(\frac{1}{12}\right)^2 = \frac{1}{144}$. Add and subtract $\frac{1}{144}$ inside the parentheses: $$12\left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| wenxinyiyan | 首先，我们需要将二次项的系数化为1，这样我们才能更容易地完成平方。我们可以通过除以12来实现这一点： $12x^2 + 2x + 11 = 12\left(x^2 + \frac{1}{6}x + \frac{11}{12}\right)$ 接下来，我们需要找到一个数，使得$x^2 + \frac{1}{6}x$可以变成一个完全平方。这个数应该是$\frac{1}{6}$的一半，即$\frac{1}{12}$，然后平方它，得到$\left(\frac{1}{12}\right)^2 = \frac{1}{144}$。 现在，我们可以将$\frac{1}{144}$加到$x^2 + \frac{1}{6}x$中，并从$\frac{11}{12}$中减去相应的量，以保持等式的平衡： $12\left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144} + \frac{11}{12}\right)$ 现在我们可以完成平方了： $= 12\left(\left(x + \frac{1}{12}\right)^2 - \frac{1}{144} + \frac{11}{12}\right)$ 最后，我们可以将常数项合并： $= 12\left(\left(x + \frac{1}{12}\right)^2 + \frac{131}{144}\right)$ $= \boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12}}$ |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| zhipu | To complete the square for the quadratic $12x^2 + 2x + 11$, follow these steps: 1. **Factor out the leading coefficient** from the terms involving $x$: \[ 12x^2 + 2x + 11 = 12(x^2 + \frac{2}{12}x) + 11 = 12(x^2 + \frac{1}{6}x) + 11 \] 2. **Complete the square** inside the parentheses: - Take half of the coefficient of $x$, square it, and add and subtract it inside the parentheses: \[ \left(\frac{1}{6} \div 2\right)^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144} \] - Add and subtract $\frac{1}{144}$ inside the parentheses: \[ 12\left(x^2 + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + 11 \] - Simplify the expression inside the parentheses: \[ 12\left(\left(x + \frac{1}{12}\right)^2 - \frac{1}{144}\right) + 11 \] 3. **Distribute the 12** and combine the constants: \[ 12\left(x + \frac{1}{12}\right)^2 - 12 \cdot \frac{1}{144} + 11 = 12\left(x + \frac{1}{12}\right)^2 - \frac{1}{12} + 11 \] - Convert $\frac{1}{12}$ to a common denominator with 11: \[ -\frac{1}{12} + 11 = -\frac{1}{12} + \frac{132}{12} = \frac{131}{12} \] 4. **Write the final expression**: \[ 12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12} \] Thus, the completed square form of the quadratic $12x^2 + 2x + 11$ is: \[ \boxed{12\left(x + \frac{1}{12}\right)^2 + \frac{131}{12}} \] |
12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-5bdaa1da-e1e9-9876-bc48-45611737dd4a', 'request_id': '5bdaa1da-e1e9-9876-bc48-45611737dd4a'} | 12 \left(x+\frac{1}{12}\right)^2+\frac{131}{12} |