With a mix of variables and variables to the negative 1 power, one strategy is to eliminate the negative 1 powers by multiplying everything by a positive variable to the 1st power. This is known as multiplying everything by the variable to the 1st power (or x in some cases)
\(\dfrac{2}{3x} + 6x = 2\)What are are the x intercepts to the equation below? What does that mean? Please solve by multiplying everything by x to the 1st power.
\(\dfrac{2x^{-1}}{3} + 6x = 2\)
\((x)\dfrac{2x^{-1}}{3} + 6x(x) = 2(x)\)
\(\dfrac{2}{3} + 6x^{2} = 2x\)
\(\dfrac{2}{3} + 6x^{2} - 2x = 0\)
\(6x^{2} - 2x + \dfrac{2}{3} = 0\)
\(\dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(\dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(6)(2/3)}}{2(6)}\)
\(\dfrac{2 \pm \sqrt{4 - 16}}{12}\)
\(\dfrac{2 \pm \sqrt{-12}}{12}\)
\(\dfrac{2 \pm \sqrt{-4 * 3}}{12}\)
\(\dfrac{2 \pm 2i \sqrt{3}}{12}\)
\(\dfrac{2}{12} + \dfrac{2i\sqrt{3}}{12} = \dfrac{1}{6} + \dfrac{i\sqrt{3}}{6}\)
\(and\)
\(\dfrac{2}{12} - \dfrac{2i\sqrt{3}}{12} = \dfrac{1}{6} - \dfrac{i\sqrt{3}}{6}\)
\(so:\)
\(x = \dfrac{1 + i\sqrt{3}}{6},\,\,x = \dfrac{1 - i \sqrt{3}}{6}\)
Imaginary numbers are in the x values, so no x intercepts on graph.
