What is |2^4x^2-1-5|>= 3 ?

The solution to |2^4x^2-1-5|>= 3 is x = 3/4

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|2^4x^2-1-5|>= 3

2^{4} x^{2} - 1 - 5 \geq 3

x \leq - \frac{3}{4} or - \frac{3}{4} \leq x \leq \frac{3}{4} or x \geq \frac{3}{4}

Solution steps

2^{4} x^{2} - 1 - 5 \geq 3

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

2^{4} x^{2} - 1 - 5 \leq - 3 or 2^{4} x^{2} - 1 - 5 \geq 3

2^{4} x^{2} - 1 - 5 \leq - 3 or 2^{4} x^{2} - 1 - 5 \geq 3

- \frac{3}{4} \leq x \leq \frac{3}{4} or (x \leq - \frac{3}{4} or x \geq \frac{3}{4})

Merge Overlapping Intervals

x \leq - \frac{3}{4} or - \frac{3}{4} \leq x \leq \frac{3}{4} or x \geq \frac{3}{4}

∣ 3 - 4353 x - 2.5 ∣ \geq 2

\frac{n}{2 - 4} > 3

∣ x - 6 ∣ < x^{2} - 5 x + 9

∣ x ∣ + ∣ x - 2 ∣ > 3

∣ x ∣ < ∣ x + 1 ∣

What is |2^4x^2-1-5|>= 3 ?
The solution to |2^4x^2-1-5|>= 3 is x<=-3/4 \lor-(sqrt(3))/4 <= x<= (sqrt(3))/4 \lor x>= 3/4