What is sqrt(2x+1)>0 ?

The solution to sqrt(2x+1)>0 is x>-1/2

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sqrt(2x+1)>0

2 x + 1 > 0

x > - \frac{1}{2}

Solution steps

2 x + 1 > 0

Square both sides

(2 x + 1)^{2} > 0^{2}

Simplify

2 x + 1 > 0

2 x + 1 > 0 : x > - \frac{1}{2}

x > - \frac{1}{2}

Find singularity points

Find non-negative values for radicals:

x \geq - \frac{1}{2}

Combine the intervals

x > - \frac{1}{2} and x \geq - \frac{1}{2}

Merge Overlapping Intervals

x > - \frac{1}{2}

\frac{9}{6} = \frac{x}{8}

5 (9 - z) + 3 = 23

2 x^{2} + x - 41 = - 15

∣ 2 x - 15 ∣ \leq 7

4 x + 10 = 2 (2 x + 5)