All zeros of polynomial function 3x^4 +14x^2 -5

Correct  x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

Step-by-step explanation:

Given:

3 x⁴ + 14 x² - 5 = 0   biquadratic equation

this equation is solved by a shift  x² = t and get:

3 t² + 14 t - 5 = 0

t₁₂ = (-14 ± √14² - 4 · 3 · 5) / 2 · 3 = (-14 ± √196 + 60) / 6

t₁₂ = (-14 ± √256) / 6 = (-14 ± 16) / 6

t₁ = -5   or  t₂ = 1 / 3

the solution t₁ = -5 is not accepted because it cannot be x² = -5

we accepted  t₂ = 1 / 3

x² =  1 / 3  ⇒  

x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

God is with you!!!

8y^50

step-by-step explanation: