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

Correct answer: 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

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