Quadratic Equations
6. Equations Reducible to Quadratic Form
6.2. Form 2
Equation of the form \(a{\left[ {f\left( {x + \frac{1}{x}} \right)} \right]^2} + b\left[ {f\left( {x + \frac{1}{x}} \right)} \right] + c = 0\). In such cases assume \(x + \frac{1}{x} = t\). Look at the following example.
Example: Solve the equations \(\left( {{x^2} + \frac{1}{{{x^2}}}} \right) - 5\left( {x + \frac{1}{x}} \right) + 8 = 0\)
Solution: Suppose \(\left( {x + \frac{1}{x}} \right) = t\), then
\(\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = t^2 - 2\)
Then the equation becomes,
\({t^2}-5t + {\rm{ }}6{\rm{ }} = {\rm{ }}0\) or \(t = 2\) or 3, this implies \(\left( {x + \frac{1}{x}} \right)\) is 2 or 3. Now take case 1: \(\left( {x + \frac{1}{x}} \right) =2\), it means \(x =1\). if \(\left( {x + \frac{1}{x}} \right) =3\), it means \(x^2 - 3x + 1 = 0\)
\(x =\frac{{3 \pm \sqrt {9 - 4} }}{2} = \frac{{3 \pm \sqrt 5 }}{2}\)