Question

if x^4 + x^2 = 11/5
then find value of

$\sqrt[3]{ \frac{ x + 1}{x- 1} } + \sqrt[3]{\frac{x - 1}{x + 1} }$
​

Answer 1

Given:

x^4 + x^2 = 11/5

To Find:

find value of

$\sqrt[3]{\dfrac{x+1}{x-1}}+ \sqrt[3]{\dfrac{x-1}{x+1}}$

Solution:

here, it is given that-

x^4 + x^2 = 11/5

take, x^2=y then,

$y^2+y=11/5\\\\5y^2+5y-11=0\\\\y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\y=\dfrac{-5\pm\sqrt{25+220} }{10}\\\\y=\dfrac{-5\pm\sqrt{245} }{10}$

$\Rightarrow x^2=\dfrac{-5\pm\sqrt{245} }{10}\\\\10x^2+5=\pm\sqrt{245}=\pm15.65 \\\\10x^2= 10.65 \ or \ 10x^2=-20.65\\\\x^2=1.065 \ or \ x^2=-2.065\\\\x=1.031 \ or \ x = 1.43$

take, x = 1.031

therefore,

$\sqrt[3]{\dfrac{x+1}{x-1}}+\sqrt[3]{\dfrac{x-1}{x+1}}\\\\=\sqrt[3]{\dfrac{1.031+1}{1.031-1}}+\sqrt[3]{\dfrac{1.031-1}{1.031+1}}\\\\=\sqrt[3]{\dfrac{2.031}{0.031}}+\sqrt[3]{\dfrac{0.031}{2.031}}\\\\=\sqrt[3]{6.55}+\sqrt[3]{0.015} \\\\=1.87+0.24\\\\=2.11$

Answer 2

Step-by-step explanation:

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