Question
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ound to two decimal places \[ x(6 x+1)=2 \]

Ask by Conner Alexander. in South Africa
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions are \( x = 0.50 \) and \( x = -0.67 \).

Solution

To solve the equation \( x(6x + 1) = 2 \) and round the solutions to two decimal places, follow these steps: 1. **Expand the Equation:** \[ x(6x + 1) = 2 \\ 6x^2 + x = 2 \\ 6x^2 + x - 2 = 0 \] 2. **Identify the Coefficients:** The quadratic equation is in the form \( ax^2 + bx + c = 0 \), where: - \( a = 6 \) - \( b = 1 \) - \( c = -2 \) 3. **Use the Quadratic Formula:** The solutions for \( x \) are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in the values: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(6)(-2)}}{2(6)} \\ x = \frac{-1 \pm \sqrt{1 + 48}}{12} \\ x = \frac{-1 \pm \sqrt{49}}{12} \\ x = \frac{-1 \pm 7}{12} \] 4. **Calculate the Solutions:** - **First Solution:** \[ x = \frac{-1 + 7}{12} = \frac{6}{12} = 0.5 \] Rounded to two decimal places: \( x = 0.50 \) - **Second Solution:** \[ x = \frac{-1 - 7}{12} = \frac{-8}{12} = -0.6666\ldots \] Rounded to two decimal places: \( x = -0.67 \) 5. **Final Answer:** The solutions to the equation \( x(6x + 1) = 2 \) rounded to two decimal places are: \[ x = 0.50 \quad \text{and} \quad x = -0.67 \]

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Mind Expander

To solve the equation \( x(6x + 1) = 2 \), start by expanding it to get \( 6x^2 + x - 2 = 0 \). Now, you can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6, b = 1, c = -2 \). Plugging in these values results in \( x = \frac{-1 \pm \sqrt{1^2 - 4(6)(-2)}}{2(6)} \), which simplifies to \( x = \frac{-1 \pm \sqrt{1 + 48}}{12} \) or \( x = \frac{-1 \pm \sqrt{49}}{12} \). This gives you \( x = \frac{-1 + 7}{12} = \frac{6}{12} = 0.5 \) and \( x = \frac{-1 - 7}{12} = \frac{-8}{12} = -\frac{2}{3} \). The solutions rounded to two decimal places are \( 0.50 \) and \( -0.67 \). Additionally, double-check your work by substituting the values back into the original equation. For \( x = 0.5 \): \( 0.5(6(0.5) + 1) = 0.5(3 + 1) = 0.5 \cdot 4 = 2 \) (checks out!). For \( x = -\frac{2}{3} \): \( -\frac{2}{3}(6(-\frac{2}{3}) + 1) = -\frac{2}{3}(-4 + 1) = -\frac{2}{3} \cdot -3 = 2 \) (also checks out!). It's great to see both solutions validate the original equation!

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