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(a + b)^2 Expansion and Visualisation

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Question

Solve for $x : ∣ ∣ ∣ \frac{x^{2} + 6}{5 x} ∣ ∣ ∣ \geq 1$

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Solution

$∣ ∣ ∣ \frac{x^{2} + 6}{5 x} ∣ ∣ ∣ \geq 1$

$\frac{x^{2} + 6}{5 x} \leq - 1 o r \frac{x^{2} + 6}{5 x} \geq 1$
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$\frac{x^{2} + 6}{5 x} \leq - 1$

$\frac{x^{2} + 6}{5 x} + 1 \leq 0$

$\frac{x^{2} + 6 + 5 x}{5 x} \leq 0$

$\frac{(x + 2) (x + 3)}{5 x} \leq 0$
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$\frac{x^{2} + 6}{5 x} \geq 1$

$\frac{x^{2} + 6}{5 x} - 1 \geq 0$

$\frac{x^{2} + 6 - 5 x}{5 x} \geq 0$

$\frac{(x - 2) (x - 3)}{5 x} \geq 0$
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$x \leq 3, - 2 \leq x < 0,, 0 < x \leq 2, x \geq 3$

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