What is -2<sqrt(-x^2+9/4)-1/2 <1 ?

The solution to -2<sqrt(-x^2+9/4)-1/2 <1 is -3/2 <= x<0\lor 0<x<= 3/2

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-2<sqrt(-x^2+9/4)-1/2 <1

- 2 < - x^{2} + \frac{9}{4} - \frac{1}{2} < 1

- \frac{3}{2} \leq x < 0 or 0 < x \leq \frac{3}{2}

Interval Notation

[- \frac{3}{2}, 0) \cup (0, \frac{3}{2}]

Decimal Notation

- 1.5 \leq x < 0 or 0 < x \leq 1.5

Solution steps

- 2 < - x^{2} + \frac{9}{4} - \frac{1}{2} < 1

a < u < b

then

a < u and u < b

- 2 < - x^{2} + \frac{9}{4} - \frac{1}{2} and - x^{2} + \frac{9}{4} - \frac{1}{2} < 1

- 2 < - x^{2} + \frac{9}{4} - \frac{1}{2} : - \frac{3}{2} \leq x \leq \frac{3}{2}

- x^{2} + \frac{9}{4} - \frac{1}{2} < 1 : - \frac{3}{2} \leq x < 0 or 0 < x \leq \frac{3}{2}

Combine the intervals

- \frac{3}{2} \leq x \leq \frac{3}{2} and (- \frac{3}{2} \leq x < 0 or 0 < x \leq \frac{3}{2})

Merge Overlapping Intervals

- \frac{3}{2} \leq x < 0 or 0 < x \leq \frac{3}{2}

- \frac{2}{5} - (- \frac{1}{3})

2 x^{2} - 6 x - 8

\frac{( x - 5 ) ^{2}}{x ^{2} - 1} \geq 0

\frac{1}{n}

t, 1 - (1 + t)^{2} - r = \frac{( p \cdot t )}{m}

What is -2<sqrt(-x^2+9/4)-1/2 <1 ?
The solution to -2<sqrt(-x^2+9/4)-1/2 <1 is -3/2 <= x<0\lor 0<x<= 3/2