MHB How Do You Solve This Complex Quadratic Equation?

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To solve the complex quadratic equation $\frac{x^2+2}{x}+\frac{8x}{x^2+2}=6$, the discussion suggests simplifying it to $x^4-6x^3+12x^2+12x+4=0$. By substituting $y = \frac{x^2+2}{x}$, the equation transforms into $y + \frac{8}{y} = 6$, which can be solved as a quadratic in y. This approach allows for finding y first, and subsequently determining the values of x. The overall consensus is that the problem can be tackled without exceeding quadratic complexity.
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please help me with this

$\frac{x^2+2}{x}+\frac{8x}{x^2+2}=6$

this is where I can get to when I simplify the the equation above,

$x^4-6x^3+12x^2+12x+4=0$
 
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paulmdrdo said:
please help me with this

$\frac{x^2+2}{x}+\frac{8x}{x^2+2}=6$

this is where I can get to when I simplify the the equation above,

$x^4-6x^3+12x^2+12x+4=0$

the above has become more complex
in case you put
$\frac{x^2+2}{x}= y$

then you get
$ y +\frac{8}{y} = 6$

you get quadratic in y then solve for y and based on it solve for x

I hope you can proceed because at no stage you get more than quadratic
 
paulmdrdo said:
please help me with this

$\frac{x^2+2}{x}+\frac{8x}{x^2+2}=6$

this is where I can get to when I simplify the the equation above,

$x^4-6x^3+12x^2+12x+4=0$

Let $ y = \frac{x^2+2}{x} $
Solve it for y
$y + \frac{8}{y} = 6 $
Then solve it for x
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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