First slide

Theory of equations

Question

The number of roots of the equation $\frac{1}{x} + \frac{1}{\sqrt{(1 - x^{2})}} = \frac{35}{12}$ is

Moderate

A

0
B

1
C

2
D

3

Solution

Let $\frac{1}{x} = u$ and $\frac{1}{\sqrt{(1 - x^{2})}} = v$ then $u + v = \frac{35}{12} and u^{2} + v^{2} = u^{2} v^{2}$
$\begin{aligned} \Rightarrow (u + v)^{2} = {(\frac{35}{12})}^{2} \\ \Rightarrow u^{2} + v^{2} + 2 uv = {(\frac{35}{12})}^{2} \\ \Rightarrow u^{2} v^{2} + 2 uv = {(\frac{35}{12})}^{2} [∵ u^{2} + v^{2} = u^{2} v^{2}] \\ \Rightarrow u^{2} v^{2} + 2 uv - {(\frac{35}{12})}^{2} = 0 \\ \Rightarrow (uv + \frac{49}{12}) (uv - \frac{25}{12}) = 0 \\ uv = - \frac{49}{12}, uv = \frac{25}{12} \end{aligned}$
Case I: If $uv = - \frac{49}{12}$ then
$\begin{aligned} \frac{1}{x} \cdot \frac{1}{\sqrt{(1 - x^{2})}} = - \frac{49}{12} \\ x^{4} - x^{2} + \frac{(12)^{2}}{(49)^{2}} = 0 \\ x = - \frac{(5 + \sqrt{73})}{14} \end{aligned}$
Case II: If $uv = \frac{25}{12}$ then
$\begin{matrix} \frac{1}{x} \cdot \frac{1}{\sqrt{(1 - x^{2})}} = \frac{25}{12} [here x > 0] \\ x^{4} - x^{2} + \frac{(12)^{2}}{(25)^{2}} = 0 \\ (x^{2} - \frac{9}{25}) (x^{2} - \frac{16}{25}) = 0 \Rightarrow x = \frac{3}{5}, \frac{4}{5} \end{matrix}$
On combining both cases,
$x = - \frac{(5 + \sqrt{73})}{14}, \frac{3}{5}, \frac{4}{5}$
Hence, number of roots = 3

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