Solve. \( 10 x^{4}-19 x^{2}+6=0 \)

To solve the equation \( 10 x^{4}-19 x^{2}+6=0 \), you can make a substitution. Let \( y = x^{2} \). The equation transforms into \( 10 y^{2} - 19 y + 6 = 0 \). Now, you can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 10 \), \( b = -19 \), and \( c = 6 \). Calculating the discriminant: \[ b^2 - 4ac = (-19)^2 - 4(10)(6) = 361 - 240 = 121 \] Now use the quadratic formula: \[ y = \frac{19 \pm \sqrt{121}}{2(10)} = \frac{19 \pm 11}{20} \] This gives us: 1. \( y = \frac{30}{20} = 1.5 \) 2. \( y = \frac{8}{20} = 0.4 \) Now, substitute back for \( x^{2} \): 1. From \( x^{2} = 1.5 \), we find \( x = \pm \sqrt{1.5} = \pm \frac{\sqrt{6}}{2} \). 2. From \( x^{2} = 0.4 \), we find \( x = \pm \sqrt{0.4} = \pm \frac{\sqrt{2}}{5} \). The solution set is \( x = \pm \frac{\sqrt{6}}{2}, \pm \frac{\sqrt{2}}{5} \).