\(6x^2 = 2x-1\).
Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula
(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).
Hope this helps,
- PM
\(6x^2 = 2x-1\).
Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula
(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).
Hope this helps,
- PM