Determine-x-x-3-x-1-1-3-2-

Question Number 218047 by dscm last updated on 27/Mar/25

D e t e r m i n e x :

\sqrt{x + 3} + \sqrt[3]{x + 1} = 2

Answered by Rasheed.Sindhi last updated on 27/Mar/25

\sqrt{x + 3} + \sqrt[3]{x + 1} = 2

\sqrt[3]{x + 1} = y \Rightarrow x = y^{3} - 1

\sqrt{y^{3} + 2} = 2 - y

y^{3} + 2 = y^{2} + 4 - 4 y

y^{3} - y^{2} + 4 y - 2 = 0

\dots

Answered by Frix last updated on 27/Mar/25

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

x + 1 = 6 x + 26 - (x + 15) \sqrt{x + 3}

(x + 15) \sqrt{x + 3} = 5 x + 25

{(x + 15)}^{2} (x + 3) = {(5 x + 25)}^{2}

x^{3} + 8 x^{2} + 65 x + 50 = 0

x = - \frac{1}{3} (8 - \sqrt[3]{1153 + 150 \sqrt{159}} + \sqrt[3]{- 1153 + 150 \sqrt{159}})

\dots better use a calculator to approximate

x \approx - . 848429802

Answered by vnm last updated on 27/Mar/25

x + 3 = u^{2}

u + \sqrt[3]{u^{2} - 2} = 2

u^{2} - 2 = {(2 - u)}^{3}

u^{2} - 2 = 8 - 12 u + 6 u^{2} - u^{3}

u^{3} - 5 u^{2} + 12 u - 10 = 0

u = t + \frac{5}{3}

t^{3} + \frac{11}{3} t + \frac{20}{27} = 0

t = \sqrt[3]{- \frac{10}{27} + \sqrt{\frac{53}{27}}} + \sqrt[3]{- \frac{10}{27} - \sqrt{\frac{53}{27}}}

x = {(t + \frac{5}{3})}^{2} - 3

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