x-2-1-3-x-3-1-3-gt-1-2-

Question Number 53108 by Tawa1 last updated on 17/Jan/19

\sqrt[3]{x + 2} - \sqrt[3]{x - 3} > \frac{1}{2}

Answered by kaivan.ahmadi last updated on 17/Jan/19

t^{3} = x - 3 \Rightarrow t^{3} + 5 = x + 2 \Rightarrow

\sqrt[3]{t^{3} + 5} ⟩ t + \frac{1}{2} \Rightarrow power 3

t^{3} + 5 ⟩ t^{3} + \frac{3}{2} t^{2} + \frac{3}{4} t + \frac{1}{8} \Rightarrow

{12 t}^{2} + 6 t - 39 ⟨ 0 \Rightarrow {4 t}^{2} + 2 t - 13 ⟨ 0 \Rightarrow

Δ = 4 + 208 = 212

t_{1, 2} = \frac{- 2 \pm \sqrt{212}}{8} = \frac{- 1 \pm \sqrt{53}}{4}

x_{1, 2} = t^{3} + 3 = {(\frac{- 1 \pm \sqrt{53}}{4})}^{3} + 3

now we can solve easily

Commented by Tawa1 last updated on 18/Jan/19

God bless you sir

Facebook Tweet Pin