First slide

Theory of equations

Question

The equation ${(\frac{x}{x + 1})}^{2} + {(\frac{x}{x - 1})}^{2} = a (a - 1)$ has

Moderate

A

four real roots if a > 2
B

four real roots if a < - 1
C

two real roots if 1 < a < 2
D

no real root if a < - 1

Solution

$\begin{array}{l} {(\frac{x}{x + 1})}^{2} + {(\frac{x}{x - 1})}^{2} = a (a - 1) \\ or {(\frac{x}{x + 1} + \frac{x}{x - 1})}^{2} - 2 (\frac{x}{x + 1}) (\frac{x}{x - 1}) = a (a - 1) \\ or {(\frac{2 x^{2}}{x^{2} - 1})}^{2} - \frac{2 x^{2}}{x^{2} - 1} - a (a - 1) = 0 \\ or t^{2} - t - a (a - 1) = 0 where t = \frac{2 x^{2}}{x^{2} - 1} \\ \Rightarrow t = a or 1 - a \\ \Rightarrow \frac{2 x^{2}}{x^{2} - 1} = a or \frac{2 x^{2}}{x^{2} - 1} = 1 - a \\ \Rightarrow x = \pm \sqrt{\frac{a}{a - 2}} or x = \pm \sqrt{\frac{a - 1}{a + 1}} \end{array}$
When $a < - 1 \Rightarrow$ all roots are real
$1 < a < 2 \Rightarrow x = \pm \sqrt{\frac{a}{2 - a}} i, \pm \sqrt{\frac{a - 1}{a + 1}}$
$\Rightarrow$ two real roots
When $a > 2 \Rightarrow$ all roots are real

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