What is |((x^2+12))/((7x))|>= 1 ?

The solution to |((x^2+12))/((7x))|>= 1 is x = 4

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|((x^2+12))/((7x))|>= 1

\frac{( x ^{2} + 12 )}{( 7 x )} \geq 1

x \leq - 4 or - 3 \leq x < 0 or 0 < x \leq 3 or x \geq 4

Solution steps

\frac{x ^{2} + 12}{7 x} \geq 1

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

\frac{x ^{2} + 12}{7 x} \leq - 1 or \frac{x ^{2} + 12}{7 x} \geq 1

\frac{x ^{2} + 12}{7 x} \leq - 1 or \frac{x ^{2} + 12}{7 x} \geq 1

(x \leq - 4 or - 3 \leq x < 0) or (0 < x \leq 3 or x \geq 4)

Merge Overlapping Intervals

x \leq - 4 or - 3 \leq x < 0 or 0 < x \leq 3 or x \geq 4

\frac{( ∣ x ∣ - 4 )}{( ∣ x ∣ - 5 )} \leq 0

∣ x ∣ - ∣ x - 2 ∣ \geq 1

\frac{3 - x}{2} \leq 2

\frac{( 2 x - 4 )}{- 6} + 2 \leq 10

∣ z - 1 ∣ + ∣ z + 1 ∣ \leq 3

What is |((x^2+12))/((7x))|>= 1 ?
The solution to |((x^2+12))/((7x))|>= 1 is x<=-4\lor-3<= x<0\lor 0<x<= 3\lor x>= 4