How Do You Solve This Complex Quadratic Equation?

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SUMMARY

The discussion centers on solving the complex quadratic equation derived from the expression $\frac{x^2+2}{x}+\frac{8x}{x^2+2}=6$. The user simplifies this to the polynomial equation $x^4-6x^3+12x^2+12x+4=0$. A substitution is suggested where $y = \frac{x^2+2}{x}$, leading to the quadratic equation $y + \frac{8}{y} = 6$. This method allows for solving for $y$ and subsequently for $x$ without exceeding quadratic complexity.

PREREQUISITES
  • Understanding of polynomial equations and their solutions
  • Familiarity with substitution methods in algebra
  • Knowledge of quadratic equations and their properties
  • Basic skills in manipulating algebraic fractions
NEXT STEPS
  • Study the process of polynomial long division to simplify complex equations
  • Learn about the quadratic formula and its applications
  • Explore substitution techniques in algebra for solving equations
  • Investigate the properties of rational functions and their graphs
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Students, educators, and anyone interested in advanced algebraic techniques, particularly those tackling complex polynomial equations and quadratic forms.

paulmdrdo1
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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
 

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