What is |x^2-4|<= 3 ?

The solution to |x^2-4|<= 3 is -sqrt(7)<= x<=-1\lor 1<= x<= sqrt(7)

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|x^2-4|<= 3

x^{2} - 4 \leq 3

- 7 \leq x \leq - 1 or 1 \leq x \leq 7

Interval Notation

[- 7, - 1] \cup [1, 7]

Decimal Notation

- 2.64575 \dots \leq x \leq - 1 or 1 \leq x \leq 2.64575 \dots

Solution steps

x^{2} - 4 \leq 3

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 3 \leq x^{2} - 4 \leq 3

x^{2} - 4 \geq - 3 and x^{2} - 4 \leq 3

(x \leq - 1 or x \geq 1) and - 7 \leq x \leq 7

Merge Overlapping Intervals

- 7 \leq x \leq - 1 or 1 \leq x \leq 7

(6 - 4)^{2}

\frac{5}{2} x - 4 = 7

∣ 2 x - 1 ∣ = 1

7 (x - 2) = 4 x + 8

x^{2} - x + 6 > 0

What is |x^2-4|<= 3 ?
The solution to |x^2-4|<= 3 is -sqrt(7)<= x<=-1\lor 1<= x<= sqrt(7)