The number of solutions of $\frac{{(2 | x | - 3)}^{2} - | x | - 6}{4 x + 1} = 0$ which belong to the domain of definition of the function y = $\frac{(2 x + 1)}{(x - 36)}$ is

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Solution

Given function
$y = \frac{2 x + 1}{x - 36}$
Domain is $R - {36}$
Now, $\frac{{(2 | x | - 3)}^{2} - | x | - 6}{4 x + 1} = 0$
Put $| x | = t$
$\Rightarrow (2 t - 3)^{2} - t - 6 = 0$
$\Rightarrow 4 t^{2} - 13 t + 3 = 0$
$\Rightarrow (t - 3) (4 t - 1) = 0$
$\Rightarrow | x | = 3, | x | = \frac{1}{4}$
$\Rightarrow x = - 3, 3, \frac{1}{4}, - \frac{1}{4}$
But $x = - \frac{1}{4}$ is not possible
Number of solutions =3

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\frac{(2 | x | - 3)^{2} - | x | - 6}{4 x + 1} = 0

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