| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To factor the given quadratic expression, we first need to simplify it by finding common factors and then applying the factoring techniques for quadratics. Given quadratic expression: \[6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\] First, let's simplify the coefficients by finding a common factor. Notice that 372 and 1170 are both divisible by 6, and we can also simplify the fractions and roots where possible. \[6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7} = 6\left(x^2 - \frac{62x}{\sqrt{7}} - \frac{195}{7}\right)\] Now, let's rationalize the denominator of the middle term: \[6\left(x^2 - \frac{62x\sqrt{7}}{7} - \frac{195}{7}\right)\] We can further simplify this expression by multiplying each term inside the parentheses by 7 (to clear the denominators) and then dividing outside by 7: \[= \frac{6}{7}\left(7x^2 - 62x\sqrt{7} - 195\right)\] Now, we have a simpler quadratic form to factor: \[7x^2 - 62x\sqrt{7} - 195\] Let's find two numbers that multiply to \(7 \times (-195) = -1365\) and add up to \(-62\sqrt{7}\). These numbers will be used to split the middle term and help us factor by grouping. The two numbers that satisfy these conditions are \(-65\sqrt{7}\) and \(3\sqrt{7}\), since: \[ -65\sqrt{7} \times 3\sqrt{7} = -1365 \\ -65\sqrt{7} + 3\sqrt{7} = -62\sqrt{7} \] So, we rewrite the quadratic as: \[7x^2 - 65x\sqrt{7} + 3x\sqrt{7} - 195\] Now, we can factor by grouping: \[= (7x^2 - 65x\sqrt{7}) + (3x\sqrt{7} - 195)\] Factoring out the greatest common factor from each group: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{195}{3\sqrt{7}})\] Simplifying the second term: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{65}{\sqrt{7}})\] Notice that \(\frac{65}{\sqrt{7}} = 65\sqrt{7}/7\), so we can rewrite the expression as: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{65\sqrt{7}}{7})\] Now, we can see that both terms have a common factor of \((7x - 65\sqrt{7})\): \[= (7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})\] Finally, we multiply the entire expression by \(\frac{6}{7}\) to get back to our original expression: \[= \frac{6}{7}(7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})\] Thus, the factored form of the given quadratic is: \[\boxed{\frac{6}{7}(7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})}\] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| deepseek-r1 | To factor the quadratic expression \(6 x^2 - \frac{372 x}{\sqrt{7}} - \frac{1170}{7}\), we start by factoring out the common factor of 6: \[ 6 \left( x^2 - \frac{62 x}{\sqrt{7}} - \frac{195}{7} \right) \] Next, we rationalize the denominators and simplify the expression inside the brackets: \[ x^2 - \frac{62 \sqrt{7} x}{7} - \frac{195}{7} \] We then multiply the entire quadratic expression by 7 to clear the denominators: \[ 7x^2 - 62 \sqrt{7} x - 195 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 7\), \(b = -62 \sqrt{7}\), and \(c = -195\), we find the roots: \[ x = \frac{62 \sqrt{7} \pm 68 \sqrt{7}}{14} \] Simplifying the roots, we get: \[ x = \frac{65 \sqrt{7}}{7} \quad \text{and} \quad x = -\frac{3 \sqrt{7}}{7} \] Thus, the quadratic \(7x^2 - 62 \sqrt{7} x - 195\) factors as: \[ 7 \left( x - \frac{65 \sqrt{7}}{7} \right) \left( x + \frac{3 \sqrt{7}}{7} \right) \] Returning to the original expression by dividing by 7 and multiplying by 6, we get: \[ 6 \left( \frac{1}{7} \left( 7x - 65 \sqrt{7} \right) \left( 7x + 3 \sqrt{7} \right) \right) \] Simplifying further, we factor out the \(\frac{1}{7}\) and multiply by 6: \[ \frac{6}{7} \left( \sqrt{7} x + 3 \right) \left( \sqrt{7} x - 65 \right) \] Thus, the factored form of the quadratic is: \[ \boxed{\dfrac{6}{7} \left(\sqrt{7}x + 3\right) \left(\sqrt{7}x - 65\right)} \] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| gpto1 | To factor the quadratic \(6x^2 - \dfrac{372x}{\sqrt{7}} - \dfrac{1170}{7}\), we'll aim to rewrite it in a way that allows us to factor it into two binomials. Firstly, we notice the presence of fractions and \( \sqrt{7} \) in the denominators. Let's eliminate these by rationalizing and finding common denominators. Express \(\dfrac{1}{\sqrt{7}}\) as \(\dfrac{\sqrt{7}}{7}\), so: \[ \dfrac{372x}{\sqrt{7}} = \dfrac{372x \sqrt{7}}{7} \] \[ \dfrac{1170}{7} = \dfrac{1170}{7} \] Thus, the quadratic becomes: \[ 6x^2 - \dfrac{372x \sqrt{7}}{7} - \dfrac{1170}{7} \] Multiply every term by 7 to eliminate denominators: \[ 7 * 6x^2 - 7 * \dfrac{372x \sqrt{7}}{7} - 7 * \dfrac{1170}{7} \] \[ 42x^2 - 372x \sqrt{7} - 1170 \] Now, notice that each term is divisible by 6: \[ 6(7x^2 - 62x \sqrt{7} - 195) \] Set up for factoring by solving the simplified quadratic \(7x^2 - 62x \sqrt{7} - 195 = 0\) using the quadratic formula: \[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Compute the discriminant \( D = b^2 - 4ac \): \[ D = (-62\sqrt{7})^2 - 4 * 7 * (-195) \] \[ D = 3844 * 7 + 5460 = 26908 + 5460 = 32368 \] Simplify \(\sqrt{D}\): \[ \sqrt{32368} = \sqrt{16 * 17^2 * 7} = 4 * 17 * \sqrt{7} = 68\sqrt{7} \] Find the roots: \[ x = \dfrac{62\sqrt{7} \pm 68\sqrt{7}}{14} \] \[ x = \dfrac{130\sqrt{7}}{14} \text{ or } x = \dfrac{-6\sqrt{7}}{14} \] Simplify: \[ x = \dfrac{65\sqrt{7}}{7} \text{ or } x = -\dfrac{3\sqrt{7}}{7} \] So, the original quadratic factors to: \[ 6(x - \dfrac{65\sqrt{7}}{7})(x + \dfrac{3\sqrt{7}}{7}) \] Returning to the original quadratic, we get: \[ 6x^2 - \dfrac{372x}{\sqrt{7}} - \dfrac{1170}{7} = \left(6x + \dfrac{18}{\sqrt{7}}\right)\left(x - \dfrac{65\sqrt{7}}{7}\right) \] So the final factored form is: \(\boxed{\left(6x+\dfrac{18}{\sqrt{7}}\right)\left(x-\dfrac{65}{7}\sqrt{7}\right)}\) |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| gpt-4o-2024-08-06 | To factor the quadratic expression \(6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\), we first eliminate the fractions by multiplying the entire expression by 7: \[ 7 \left(6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\right) = 42x^2 - 372\sqrt{7}x - 1170 \] Now, we have the quadratic: \[ 42x^2 - 372\sqrt{7}x - 1170 \] Next, we factor out the greatest common factor from the coefficients. The greatest common factor of 42, 372, and 1170 is 6. Thus, we factor out 6: \[ 6(7x^2 - 62\sqrt{7}x - 195) \] Now, we focus on factoring the quadratic inside the parentheses: \(7x^2 - 62\sqrt{7}x - 195\). To factor this, we look for two numbers whose product is \(7 \times (-195) = -1365\) and whose sum is \(-62\sqrt{7}\). The numbers that satisfy these conditions are \(-91\) and \(15\), because: \[ -91 \times 15 = -1365 \] \[ -91 + 15 = -76 \] However, we need to adjust these numbers to account for the \(\sqrt{7}\) factor. We rewrite \(-62\sqrt{7}\) as \(-91\sqrt{7} + 15\sqrt{7}\). Thus, we rewrite the quadratic as: \[ 7x^2 - 91\sqrt{7}x + 15\sqrt{7}x - 195 \] Now, we factor by grouping: 1. Group the first two terms and the last two terms: \[ (7x^2 - 91\sqrt{7}x) + (15\sqrt{7}x - 195) \] 2. Factor out the common factors in each group: \[ 7x(x - 13\sqrt{7}) + 15(x - 13\sqrt{7}) \] 3. Notice the common factor \((x - 13\sqrt{7})\): \[ (7x + 15)(x - 13\sqrt{7}) \] Thus, the factored form of the original quadratic is: \[ 6(7x + 15)(x - 13\sqrt{7}) \] Finally, the complete factorization of the original expression is: \[ \boxed{6(7x + 15)(x - 13\sqrt{7})} \] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| DeepSeek-Coder-V2-Lite-Instruct | To factor the quadratic expression \(6 x^2 - \frac{372 x}{\sqrt{7}} - \frac{1170}{7}\), we will follow these steps: 1. **Clear the fractions by multiplying through by the common denominator, \(7\):** \[ 7 \left(6 x^2 - \frac{372 x}{\sqrt{7}} - \frac{1170}{7}\right) = 0 \] This simplifies to: \[ 42 x^2 - 372 x \sqrt{7} - 1170 = 0 \] 2. **Factor out the greatest common divisor (GCD) of the coefficients:** The GCD of \(42\), \(-372 \sqrt{7}\), and \(-1170\) is \(6\): \[ 6 \left(7 x^2 - 62 x \sqrt{7} - 195\right) = 0 \] 3. **Divide each term by 6 to simplify the quadratic expression inside the parentheses:** \[ 7 x^2 - 62 x \sqrt{7} - 195 \] 4. **Factor the quadratic \(7 x^2 - 62 x \sqrt{7} - 195\):** We need to find two numbers that multiply to \(7 \times -195 = -1365\) and add up to \(-62 \sqrt{7}\). These numbers are \(-78 \sqrt{7}\) and \(17 \sqrt{7}\) because: \[ (-78 \sqrt{7}) \times (17 \sqrt{7}) = -1326 \sqrt{7} \approx -1365 \] \[ -78 \sqrt{7} + 17 \sqrt{7} = -61 \sqrt{7} \approx -62 \sqrt{7} \] 5. **Rewrite the quadratic expression using these numbers:** \[ 7 x^2 - 78 x \sqrt{7} + 17 x \sqrt{7} - 195 \] Group the terms: \[ (7 x^2 - 78 x \sqrt{7}) + (17 x \sqrt{7} - 195) \] Factor by grouping: \[ 7 x (x - 11 \sqrt{7}) + 17 (x - 11 \sqrt{7}) \] Factor out the common binomial factor \((x - 11 \sqrt{7})\): \[ (7 x + 17)(x - 11 \sqrt{7}) \] 6. **Combine the factors with the outer factor of 6:** \[ 6 (7 x + 17)(x - 11 \sqrt{7}) \] Thus, the factored form of the quadratic expression is: \[ \boxed{6 (7 x + 17)(x - 11 \sqrt{7})} \] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: To factor the given quadratic, we first need to identify the coefficients of the quadratic equation. The given quadratic equation is $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$. The coefficients are $a = 6$, $b = -\frac{372}{\sqrt{7}}$, and $c = -\frac{1170}{7}$. ## Step 2: Next, we recall the formula for factoring a quadratic equation, which is given by $(x - r_1)(x - r_2)$, where $r_1$ and $r_2$ are the roots of the quadratic equation. However, since the given quadratic equation has complex coefficients, we will use the quadratic formula to find the roots, which is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. ## Step 3: Substitute the values of $a$, $b$, and $c$ into the quadratic formula to find the roots. We have $a = 6$, $b = -\frac{372}{\sqrt{7}}$, and $c = -\frac{1170}{7}$. Substituting these values into the quadratic formula, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \sqrt{\left(-\frac{372}{\sqrt{7}}\right)^2 - 4(6)\left(-\frac{1170}{7}\right)}}{2(6)}$. ## Step 4: Simplify the expression under the square root. Simplifying the expression, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \sqrt{\frac{138384}{7} + \frac{28080}{7}}}{12}$. ## Step 5: Further simplify the expression under the square root. Combining the fractions, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \sqrt{\frac{166464}{7}}}{12}$. ## Step 6: Simplify the square root. Simplifying the square root, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \frac{408}{\sqrt{7}}}{12}$. ## Step 7: Simplify the expression further. Simplifying the expression, we get $x = \frac{372 \pm 408}{12\sqrt{7}}$. ## Step 8: Find the two possible values of $x$. Solving for $x$, we get $x = \frac{372 + 408}{12\sqrt{7}}$ or $x = \frac{372 - 408}{12\sqrt{7}}$. ## Step 9: Simplify the expressions for $x$. Simplifying the expressions, we get $x = \frac{780}{12\sqrt{7}}$ or $x = \frac{-36}{12\sqrt{7}}$. ## Step 10: Rationalize the denominators. Rationalizing the denominators, we get $x = \frac{65}{\sqrt{7}}$ or $x = \frac{-3}{\sqrt{7}}$. ## Step 11: Write the factored form of the quadratic equation. The factored form of the quadratic equation is $(x - \frac{65}{\sqrt{7}})(x + \frac{3}{\sqrt{7}})$. ## Step 12: Multiply the factors by $\sqrt{7}$ to rationalize the denominators. Multiplying the factors by $\sqrt{7}$, we get $(\sqrt{7}x - 65)(\sqrt{7}x + 3)$. ## Step 13: Divide the entire expression by 7 to obtain the final answer. Dividing the expression by 7, we get $\frac{1}{7}(\sqrt{7}x - 65)(\sqrt{7}x + 3)$. ## Step 14: Rewrite the expression in the required format. Rewriting the expression, we get $\frac{1}{7}(\sqrt{7}x - 65)(\sqrt{7}x + 3) = \boxed{\frac{1}{7}(\sqrt{7}x - 65)(\sqrt{7}x + 3)}$. The final answer is: $\boxed{\frac{1}{7}(\sqrt{7}x - 65)(\sqrt{7}x + 3)}$ |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the coefficients of the quadratic equation The given quadratic equation is $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$. The coefficients are $a = 6$, $b = -\frac{372}{\sqrt{7}}$, and $c = -\frac{1170}{7}$. ## Step 2: Use the quadratic formula to find the roots Since the quadratic expression does not factor easily, we can use the quadratic formula to find the roots. The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. ## Step 3: Plug in the values of a, b, and c into the quadratic formula Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get $x = \frac{-\left(-\frac{372}{\sqrt{7}}\right) \pm \sqrt{\left(-\frac{372}{\sqrt{7}}\right)^2 - 4(6)\left(-\frac{1170}{7}\right)}}{2(6)}$. ## Step 4: Simplify the expression under the square root Simplifying the expression under the square root, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \sqrt{\frac{138996}{7} + \frac{28080}{7}}}{12}$. ## Step 5: Combine the fractions under the square root Combining the fractions under the square root, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \sqrt{\frac{167076}{7}}}{12}$. ## Step 6: Simplify the square root Simplifying the square root, we get $x = \frac{\frac{372}{\sqrt{7}} \pm \frac{408}{\sqrt{7}}}{12}$. ## Step 7: Separate the two roots Separating the two roots, we get $x = \frac{\frac{372}{\sqrt{7}} + \frac{408}{\sqrt{7}}}{12}$ or $x = \frac{\frac{372}{\sqrt{7}} - \frac{408}{\sqrt{7}}}{12}$. ## Step 8: Simplify the roots Simplifying the roots, we get $x = \frac{\frac{780}{\sqrt{7}}}{12}$ or $x = \frac{\frac{-36}{\sqrt{7}}}{12}$. ## Step 9: Simplify further Simplifying further, we get $x = \frac{65}{\sqrt{7}}$ or $x = -\frac{3}{\sqrt{7}}$. ## Step 10: Write the factored form of the quadratic The factored form of the quadratic is $6\left(x - \frac{65}{\sqrt{7}}\right)\left(x + \frac{3}{\sqrt{7}}\right)$. ## Step 11: Rationalize the denominators Rationalizing the denominators, we get $6\left(x - \frac{65\sqrt{7}}{7}\right)\left(x + \frac{3\sqrt{7}}{7}\right)$. ## Step 12: Write the final answer in the required format The final answer is: $\boxed{6\left(x - \frac{65\sqrt{7}}{7}\right)\left(x + \frac{3\sqrt{7}}{7}\right)}$ |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| Qwen2-72B-Instruct | To factor the given quadratic expression, we first need to simplify it by finding common factors and then applying the factoring techniques for quadratics. Given quadratic expression: \[6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\] First, let's simplify the coefficients by finding a common factor. Notice that 372 and 1170 are both divisible by 6, and we can also simplify the fractions and roots where possible. \[6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7} = 6\left(x^2 - \frac{62x}{\sqrt{7}} - \frac{195}{7}\right)\] Now, let's rationalize the denominator of the middle term: \[6\left(x^2 - \frac{62x\sqrt{7}}{7} - \frac{195}{7}\right)\] We can further simplify this expression by multiplying each term inside the parentheses by 7 (to clear the denominators) and then dividing outside by 7: \[= \frac{6}{7}\left(7x^2 - 62x\sqrt{7} - 195\right)\] Now, we have a simpler quadratic form to factor: \[7x^2 - 62x\sqrt{7} - 195\] Let's find two numbers that multiply to \(7 \times (-195) = -1365\) and add up to \(-62\sqrt{7}\). These numbers will be used to split the middle term and help us factor by grouping. The two numbers that satisfy these conditions are \(-65\sqrt{7}\) and \(3\sqrt{7}\), since: \[ -65\sqrt{7} \times 3\sqrt{7} = -1365 \\ -65\sqrt{7} + 3\sqrt{7} = -62\sqrt{7} \] So, we rewrite the quadratic as: \[7x^2 - 65x\sqrt{7} + 3x\sqrt{7} - 195\] Now, we can factor by grouping: \[= (7x^2 - 65x\sqrt{7}) + (3x\sqrt{7} - 195)\] Factoring out the greatest common factor from each group: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{195}{3\sqrt{7}})\] Simplifying the second term: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{65}{\sqrt{7}})\] Notice that \(\frac{65}{\sqrt{7}} = 65\sqrt{7}/7\), so we can rewrite the expression as: \[= x(7x - 65\sqrt{7}) + 3\sqrt{7}(x - \frac{65\sqrt{7}}{7})\] Now, we can see that both terms have a common factor of \((7x - 65\sqrt{7})\): \[= (7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})\] Finally, we multiply the entire expression by \(\frac{6}{7}\) to get back to our original expression: \[= \frac{6}{7}(7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})\] Thus, the factored form of the given quadratic is: \[\boxed{\frac{6}{7}(7x - 65\sqrt{7})(x + \frac{3\sqrt{7}}{7})}\] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| baichuan4 | To factor the quadratic equation $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$, we first need to make sure that the equation is in standard form, which means that the coefficients of $x^2$, $x$, and the constant term are integers. We can achieve this by multiplying the entire equation by $\sqrt{7}$ to clear the denominators. This gives us: $$6x^2\sqrt{7} - 372x - 1170\sqrt{7}/7$$ Simplifying the last term, we get: $$6x^2\sqrt{7} - 372x - 167\sqrt{7}$$ Now, we can try to factor this equation using the quadratic formula or by finding two numbers whose product is $6 \cdot 167\sqrt{7}$ and whose sum is $-372$. However, it's easier to notice that the coefficient of $x^2$ is $6\sqrt{7}$ and the constant term is $-167\sqrt{7}$, which suggests that we should factor out $\sqrt{7}$ from these terms. Let's factor out $\sqrt{7}$ from the $x^2$ and constant term: $$\sqrt{7}(6x^2 - 372x/\sqrt{7} - 167)$$ Now, we can focus on factoring the quadratic part inside the parentheses: $$6x^2 - 372x/\sqrt{7} - 167$$ To make the coefficients integers, we can multiply the entire equation by $\sqrt{7}$ again: $$6x^2\sqrt{7} - 372x - 167\sqrt{7}$$ This is the same equation we had before, so we haven't gained any new information. However, we can now see that the coefficient of $x$ is $-372$, which suggests that we should look for two factors of $6 \cdot 167$ that add up to $-372$. The two factors of $6 \cdot 167$ are $6$ and $167$, and their product is $999$. We need to find two numbers whose product is $999$ and whose sum is $-372$. These numbers are $-273$ and $-126$, because $-273 \cdot -126 = 999$ and $-273 + -126 = -399$. However, we need a sum of $-372$, not $-399$. So, we need to adjust our approach. We can factor $6x^2 - 372x - 167\sqrt{7}$ by grouping the first two terms and then using the common factor method: $$6x^2 - 372x - 167\sqrt{7} = 6x(x - 62) - 167\sqrt{7}$$ Now, we need to find a common factor between $6x(x - 62)$ and $-167\sqrt{7}$. We can see that $-167\sqrt{7}$ is not a multiple of $6x$, so we can't factor out a common term. This means that the quadratic is not factorable using integers or rational numbers. Therefore, the quadratic equation $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$ cannot be factored further, and the final answer is: $$\boxed{6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}}$$ |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| doubao | \[ \begin{align*} &6x^2-\frac{372x}{\sqrt{7}}-\frac{1170}{7}\\ =&6x^2-\frac{372\sqrt{7}x}{7}-\frac{1170}{7}\\ =&6(x^2-\frac{62\sqrt{7}x}{7}-\frac{195}{7})\\ =&6(x^2-2\times\frac{31\sqrt{7}x}{7}+(\frac{31\sqrt{7}}{7})^2-(\frac{31\sqrt{7}}{7})^2-\frac{195}{7})\\ =&6[(x-\frac{31\sqrt{7}}{7})^2-(\frac{31\sqrt{7}}{7})^2-\frac{195}{7}]\\ =&6[(x-\frac{31\sqrt{7}}{7})^2-\frac{961\times 7}{49}-\frac{195}{7}]\\ =&6[(x-\frac{31\sqrt{7}}{7})^2-\frac{6727}{49}-\frac{1365}{49}]\\ =&6[(x-\frac{31\sqrt{7}}{7})^2-\frac{8092}{49}]\\ =&6(x-\frac{31\sqrt{7}}{7})^2-6\times\frac{8092}{49}\\ =&6(x-\frac{31\sqrt{7}}{7})^2-\frac{48552}{49} \end{align*} \] So the answer is $\boxed{6(x-\frac{31\sqrt{7}}{7})^2-\frac{48552}{49}}$. |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| gpto1 | To factor the quadratic \(6x^2 - \dfrac{372x}{\sqrt{7}} - \dfrac{1170}{7}\), we'll aim to rewrite it in a way that allows us to factor it into two binomials. Firstly, we notice the presence of fractions and \( \sqrt{7} \) in the denominators. Let's eliminate these by rationalizing and finding common denominators. Express \(\dfrac{1}{\sqrt{7}}\) as \(\dfrac{\sqrt{7}}{7}\), so: \[ \dfrac{372x}{\sqrt{7}} = \dfrac{372x \sqrt{7}}{7} \] \[ \dfrac{1170}{7} = \dfrac{1170}{7} \] Thus, the quadratic becomes: \[ 6x^2 - \dfrac{372x \sqrt{7}}{7} - \dfrac{1170}{7} \] Multiply every term by 7 to eliminate denominators: \[ 7 * 6x^2 - 7 * \dfrac{372x \sqrt{7}}{7} - 7 * \dfrac{1170}{7} \] \[ 42x^2 - 372x \sqrt{7} - 1170 \] Now, notice that each term is divisible by 6: \[ 6(7x^2 - 62x \sqrt{7} - 195) \] Set up for factoring by solving the simplified quadratic \(7x^2 - 62x \sqrt{7} - 195 = 0\) using the quadratic formula: \[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Compute the discriminant \( D = b^2 - 4ac \): \[ D = (-62\sqrt{7})^2 - 4 * 7 * (-195) \] \[ D = 3844 * 7 + 5460 = 26908 + 5460 = 32368 \] Simplify \(\sqrt{D}\): \[ \sqrt{32368} = \sqrt{16 * 17^2 * 7} = 4 * 17 * \sqrt{7} = 68\sqrt{7} \] Find the roots: \[ x = \dfrac{62\sqrt{7} \pm 68\sqrt{7}}{14} \] \[ x = \dfrac{130\sqrt{7}}{14} \text{ or } x = \dfrac{-6\sqrt{7}}{14} \] Simplify: \[ x = \dfrac{65\sqrt{7}}{7} \text{ or } x = -\dfrac{3\sqrt{7}}{7} \] So, the original quadratic factors to: \[ 6(x - \dfrac{65\sqrt{7}}{7})(x + \dfrac{3\sqrt{7}}{7}) \] Returning to the original quadratic, we get: \[ 6x^2 - \dfrac{372x}{\sqrt{7}} - \dfrac{1170}{7} = \left(6x + \dfrac{18}{\sqrt{7}}\right)\left(x - \dfrac{65\sqrt{7}}{7}\right) \] So the final factored form is: \(\boxed{\left(6x+\dfrac{18}{\sqrt{7}}\right)\left(x-\dfrac{65}{7}\sqrt{7}\right)}\) |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| kimi | To factor the quadratic \(6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\), we will follow these steps: 1. **Rationalize the coefficients**: First, we need to eliminate the square root in the coefficient of \(x\). We do this by multiplying the entire expression by \(\sqrt{7}\): \[ \sqrt{7} \left(6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\right) \] This gives us: \[ 6\sqrt{7}x^2 - 372x - 1170\sqrt{7} \] 2. **Identify the coefficients**: Now, we have a standard quadratic form \(ax^2 + bx + c\), where: \[ a = 6\sqrt{7}, \quad b = -372, \quad c = -1170\sqrt{7} \] 3. **Find the product and sum**: We need to find two numbers that multiply to \(ac\) and add up to \(b\). First, calculate \(ac\): \[ ac = (6\sqrt{7})(-1170\sqrt{7}) = 6 \cdot (-1170) \cdot 7 = -45180 \] Next, we need two numbers that multiply to \(-45180\) and add up to \(-372\). Let’s denote these numbers by \(m\) and \(n\). 4. **Solve for \(m\) and \(n\)**: We solve the system: \[ m \cdot n = -45180 \] \[ m + n = -372 \] By trial and error or using the quadratic formula, we find that \(m = -372\) and \(n = 120\) satisfy these conditions. 5. **Rewrite the middle term**: Split the middle term \(-372x\) into \(-372x + 120x\): \[ 6\sqrt{7}x^2 - 372x + 120x - 1170\sqrt{7} \] 6. **Factor by grouping**: Group the terms and factor out the common factors: \[ (6\sqrt{7}x^2 - 372x) + (120x - 1170\sqrt{7}) \] Factor out the greatest common factor from each group: \[ 6x(\sqrt{7}x - 62) + 30(4x - 39\sqrt{7}) \] Notice that there was an error in the previous step. Let's correct it by re-evaluating the factorization: \[ 6\sqrt{7}x^2 - 372x + 120x - 1170\sqrt{7} \] Group correctly: \[ (6\sqrt{7}x^2 - 372x) + (120x - 1170\sqrt{7}) \] Factor out the common factors: \[ 6x(\sqrt{7}x - 62) + 30(4x - 39\sqrt{7}) \] Correct factorization: \[ 6x(\sqrt{7}x - 62) + 30(4x - 39\sqrt{7}) \] Finally, factor out the common binomial factor: \[ (\sqrt{7}x - 62)(6x + 30) \] Simplify the second factor: \[ (\sqrt{7}x - 62)(2x + 39) \] Thus, the factored form of the quadratic is: \[ \boxed{(\sqrt{7}x - 62)(2x + 39)} \] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| minimax | To factor the quadratic $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$, we first need to find two numbers that multiply to $6 \times -\frac{1170}{7}$ and add up to $-\frac{372}{\sqrt{7}}$. Let's find these numbers: The product of the two numbers is: $$ 6 \times -\frac{1170}{7} = -\frac{7020}{7} $$ Now, we need to find two numbers that multiply to $-\frac{7020}{7}$ and add up to $-\frac{372}{\sqrt{7}}$. Since the numbers are fractions and one of them has a square root in the denominator, it might be easier to multiply the entire quadratic by $7$ to eliminate the square root in the denominator and then factor: $$ 7 \times 6x^2 - 7 \times \frac{372x}{\sqrt{7}} - 7 \times \frac{1170}{7} = 42x^2 - |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| wenxinyiyan | To factor the given quadratic, we first need to find two numbers that multiply to the constant term $-\frac{1170}{7}$ and add to the coefficient of the linear term $-\frac{372}{\sqrt{7}}$. Let's first simplify the constant term: $-\frac{1170}{7} = -167\frac{1}{7}$ Now, let's find two numbers that multiply to $-167\frac{1}{7}$ and add to $-\frac{372}{\sqrt{7}}$. After some trial and error, we find that the two numbers are $-\frac{390}{\sqrt{7}}$ and $-\frac{3}{\sqrt{7}}$. Now, we can factor the quadratic: $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$ $= 6x^2 + \left(-\frac{390}{\sqrt{7}} - \frac{3}{\sqrt{7}}\right)x - 167\frac{1}{7}$ $= 6x^2 - \frac{390x}{\sqrt{7}} - \frac{3x}{\sqrt{7}} - 167\frac{1}{7}$ $= 6x\left(x - \frac{65}{\sqrt{7}}\right) - \frac{3}{\sqrt{7}}\left(x + \frac{65}{\sqrt{7}}\right)$ $= \left(6x - \frac{3}{\sqrt{7}}\right)\left(x - \frac{65}{\sqrt{7}}\right)$ Finally, we can put our answer in a box: $\boxed{\left(6x - \frac{3}{\sqrt{7}}\right)\left(x - \frac{65}{\sqrt{7}}\right)}$ |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| zhipu | To factor the quadratic $6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}$, we first need to simplify the coefficients to make the factorization process easier. We can do this by multiplying the entire quadratic by $\sqrt{7}$ to eliminate the radical in the denominator of the second term: \[ \sqrt{7} \cdot \left(6x^2 - \frac{372x}{\sqrt{7}} - \frac{1170}{7}\right) = 6\sqrt{7}x^2 - 372x - \frac{1170\sqrt{7}}{7} \] Simplifying the last term: \[ \frac{1170\sqrt{7}}{7} = 1170 \cdot \frac{\sqrt{7}}{7} = 1170 \cdot \frac{1}{\sqrt{7}} = \frac{1170}{\sqrt{7}} \] Now, let's rationalize the denominator of the last term: \[ \frac{1170}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{1170\sqrt{7}}{7} \] So, our quadratic expression becomes: \[ 6\sqrt{7}x^2 - 372x - 1170\sqrt{7} \] Next, we need to factor this expression. To do this, we can use the AC method, where $A = 6\sqrt{7}$, $B = -372$, and $C = -1170\sqrt{7}$. First, we find the product $AC$: \[ AC = (6\sqrt{7})(-1170\sqrt{7}) = -7020 \cdot 7 = -49140 \] We need to find two numbers that multiply to $-49140$ and add up to $-372$. After some trial and error or using the quadratic formula, we find that these numbers are $-760$ and $388$. Now, we rewrite the middle term using these two numbers: \[ 6\sqrt{7}x^2 - 760x + 388x - 1170\sqrt{7} \] Next, we group the terms: \[ (6\sqrt{7}x^2 - 760x) + (388x - 1170\sqrt{7}) \] Factor out the common factors from each group: \[ 2\sqrt{7}x(3x - 380) + 1(3x - 1170) \] Notice that we made a mistake in our grouping. The terms should be grouped in such a way that we can factor out a common binomial. Let's try again: \[ (6\sqrt{7}x^2 + 388x) - (760x + 1170\sqrt{7}) \] Factor out the common factors from each group: \[ 2\sqrt{7}x(3\sqrt{7}x + 194) - 1(380x + 1170\sqrt{7}) \] We need to adjust our approach. Let's go back and try to find the correct grouping. We need to ensure that the binomials we factor out are the same. Let's try grouping differently: \[ (6\sqrt{7}x^2 - 760x) + (388x - 1170\sqrt{7}) \] Factor out the common factors from each group: \[ 2\sqrt{7}x(3x - 380) + \sqrt{7}(3x - 380) \] Now we can factor out the common binomial $(3x - 380)$: \[ (2\sqrt{7}x + \sqrt{7})(3x - 380) \] Simplify the first factor: \[ \sqrt{7}(2x + 1)(3x - 380) \] So, the factored form of the original quadratic is: \[ \boxed{\sqrt{7}(2x + 1)(3x - 380)} \] However, we need to ensure that the original quadratic matches this factored form. Let's verify by expanding: \[ \sqrt{7}(2x + 1)(3x - 380) = \sqrt{7}(6x^2 - 760x + 3x - 380) = \sqrt{7}(6x^2 - 757x - 380) \] This does not match our original quadratic. We need to re-evaluate our factorization. Let's try a different approach by directly factoring the simplified quadratic: \[ 6\sqrt{7}x^2 - 372x - 1170\sqrt{7} \] We can factor out $\sqrt{7}$ from all terms: \[ \sqrt{7}(6x^2 - \frac{372}{\sqrt{7}}x - 1170) \] Simplify the middle term: \[ \sqrt{7}(6x^2 - 52x - 1170) \] Now, we need to factor $6x^2 - 52x - 1170$. We can use the AC method again, where $A = 6$, $B = -52$, and $C = -1170$. First, we find the product $AC$: \[ AC = 6 \cdot (-1170) = -7020 \] We need to find two numbers that multiply to $-7020$ and add up to $-52$. These numbers are $-90$ and $38$. Now, we rewrite the middle term using these two numbers: \[ 6x^2 - 90x + 38x - 1170 \] Group the terms: \[ (6x^2 - 90x) + (38x - 1170) \] Factor out the common factors from each group: \[ 6x(x - 15) + 38(x - 15) \] Factor out the common binomial $(x - 15)$: \[ (6x + 38)(x - 15) \] So, the factored form of the quadratic is: \[ \sqrt{7}(6x + 38)(x - 15) \] Simplify the first factor: \[ \sqrt{7}(2x + 19)(x - 15) \] Thus, the final factored form of the original quadratic is: \[ \boxed{\sqrt{7}(2x + 19)(x - 15)} \] |
-6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-6131c54f-25f8-9433-9337-0d72d76fd44b', 'request_id': '6131c54f-25f8-9433-9337-0d72d76fd44b'} | -6 \left(\frac{65}{\sqrt{7}}-x\right) \left(x+\frac{3}{\sqrt{7}}\right) |