Question

Solve this question

Answer 1

$\huge \mathfrak \blue{answer : }$

GIVEN,

$| \frac{ {x}^{2} + 6 }{5x} | \leqslant 1$

EITER,

$\frac{x {}^{2} + 6 }{ 5x} \leqslant 1 \\ = > {x }^{2} + 6 \leqslant 5x \\ = > {x}^{2} - 5x + 6 \leqslant 0 \\ = > {x}^{2} - 3x - 2x + 6 \leqslant 0 \\ = > (x - 3)(x - 2) \leqslant 0 \\ \\ so \: \: 2 \leqslant x \leqslant 3$

OR,

$- \frac{ x {}^{2} + 6 }{ 5x} \leqslant 1 \\ = > - {x }^{2} - 6 \leqslant 5x \\ = > - {x}^{2} - 5x - 6 \leqslant 0 \\ = > {x}^{2} + 5x + 6 \geqslant 0 \\ = > (x + 3)(x + 2) \\ \\ so \: - 3\leqslant x \leqslant - 2$

SO,

X=[-3,-2]U[2,3]