What is the sum of all real roots of the equation 3x^4 + 10x^2 - 8 = 0?

To find the sum of all real roots of the equation ( 3x^4 + 10x^2 - 8 = 0 ), we can simplify the problem by making a substitution. Let ( y = x^2 ). This transforms the original equation into a quadratic equation:

[

3y^2 + 10y - 8 = 0

]

We can solve this quadratic equation using the quadratic formula:

[

y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

]

where ( a = 3 ), ( b = 10 ), and ( c = -8 ). Calculating the discriminant:

[

b^2 - 4ac = 10^2 - 4(3)(-8) = 100 + 96 = 196

]

Since the discriminant is positive, there will be two distinct real roots for ( y ). Now we can find the roots:

[

y = \frac{-10 \pm \sqrt{196}}{2 \cdot 3} = \frac{-10 \pm 14}{6}

]

Calculating the two possible values