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