solve \( 6 x^{4}+8 x^{2}=26 x^{3} \)

To solve the equation \( 6 x^{4}+8 x^{2}=26 x^{3} \), we start by rearranging it into a standard polynomial form: \[ 6 x^{4} - 26 x^{3} + 8 x^{2} = 0 \] Next, we can factor out \( 2 x^2 \): \[ 2 x^2(3 x^2 - 13 x + 4) = 0 \] This gives us one solution directly from \( 2 x^2 = 0 \), leading to \( x = 0 \). Now we need to solve the quadratic \( 3 x^2 - 13 x + 4 = 0 \) using the quadratic formula, which is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in \( a = 3 \), \( b = -13 \), and \( c = 4 \): \[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \] \[ = \frac{13 \pm \sqrt{169 - 48}}{6} \] \[ = \frac{13 \pm \sqrt{121}}{6} \] \[ = \frac{13 \pm 11}{6} \] This gives us two further solutions: \[ x = \frac{24}{6} = 4 \quad \text{and} \quad x = \frac{2}{6} = \frac{1}{3} \] Thus, the complete set of solutions is: \[ x = 0, \quad x = 4, \quad x = \frac{1}{3} \]