Question

solve for x : $4(x-\frac{1}{x})^2+8(x+\frac{1}{x}=29.)$ x≠0


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Answer 1

$\huge\overbrace{\underbrace{\boxed{\boxed{\mathfrak{Answer:}}}}}$

Solution:

$\bold{Let \:(x+\frac{1}{x})^2=y }\\\\\bold{(x+\frac{1}{x} )^2-(x-\frac{1}{x} )^2=4}\\\\\bold{y^2-(x-\frac{1}{x} )^2=4}\\\\\bold{and\:(x-\frac{1}{x} )^2=y^2-4}$

$\bold{4(x-\frac{1}{x} )^2+8(x+\frac{1}{x} )=29,}\\\\\bold{4(y^2-4)+8y=29}\\\\\bold{4y^2-16+8y=29}\\\\\bold{4y^2+8y-45=0}\\\\\bold{4y^2+18y-10y-45=0}\\\\\bold{2y(2y+9)-5(2y+9)=0}$

$\bold{(2y+9)(2y-5)=0,}\\\\\boxed{y=-\frac{9}{2}\:or\:y=\frac{5}{2} }}$

$\boxed{y=-\frac{9}{2}=x+\frac{1}{x}=-\frac{9}{2} }\\\\\bold{2x^2+9x+2=0}\\\\\boxed{x=\frac{-9+\sqrt{(9)^2-4*2*2} }{2*2}=\frac{-9+\sqrt{65} }{4} }$

$\boxed{y=\frac{5}{2}=x+\frac{1}{x}=\frac{5}{2} }\\\\\bold{2x^2-4x-x+2=0}\\\\\bold{2x(x-2)-1(x-2)=0}\\\\\bold{(x-2)(2x-1)=0}\\\\\bold{x=2}\:or\: \boxed{\frac{1}{2} }$

$\large\boxed{\mathfrak{solution=\frac{-9+\sqrt{65} }{4},2,or\frac{1}{2} }}$

Answer 2

Answer:

Let(x+

x

1

)

2

=y

(x+

x

1

)

2

−(x−

x

1

)

2

=4

y

2

−(x−

x

1

)

2

=4

and(x−

x

1

)

2

=y

2

−4

\begin{gathered}\bold{4(x-\frac{1}{x} )^2+8(x+\frac{1}{x} )=29,}\\\\\bold{4(y^2-4)+8y=29}\\\\\bold{4y^2-16+8y=29}\\\\\bold{4y^2+8y-45=0}\\\\\bold{4y^2+18y-10y-45=0}\\\\\bold{2y(2y+9)-5(2y+9)=0}\end{gathered}

4(x−

x

1

)

2

+8(x+

x

1

)=29,

4(y

2

−4)+8y=29

4y

2

−16+8y=29

4y

2

+8y−45=0

4y

2

+18y−10y−45=0

2y(2y+9)−5(2y+9)=0