First slide

Theory of equations

Question

The number of real roots of
${(\frac{x - 1}{x + 1})}^{4} - 13 {(\frac{x - 1}{x + 1})}^{2} + 36 = 0, x \neq - 1$ is

Moderate

A

0
B

2
C

3
D

4

Solution

Put ${(\frac{x - 1}{x + 1})}^{2} = y$ to obtain
$y^{2} - 13 y + 36 = 0 \Rightarrow y = 4, 9$
$∴ \frac{x - 1}{x + 1} = \pm 2, \pm 3$
Thus, the given equation has four real roots.

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