What is sqrt(2x-1)<= x ?

The solution to sqrt(2x-1) = 1/2

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sqrt(2x-1)<= x

2 x - 1 \leq x

x \geq \frac{1}{2}

Interval Notation

[\frac{1}{2}, \infty)

Decimal Notation

x \geq 0.5

Solution steps

2 x - 1 \leq x

f (x) \leq g (x)

then

g (x) \geq 0

x \geq 0

2 x - 1 \leq x :

True for all

x

Find singularity points

Find non-negative values for radicals:

x \geq \frac{1}{2}

Combine the intervals

x \geq 0 and True for all x and x \geq \frac{1}{2}

Merge Overlapping Intervals

x \geq \frac{1}{2}

Verify Solutions:

x \geq \frac{1}{2}

True

The solution is

x \geq \frac{1}{2}

∣ 3 x + 1 ∣ + 10 = 4

1.8 (\frac{3}{2}) 0.08341 (x - 700) = - 2 (15 - 7.5)

2 k^{4} + 4 k^{3} - 70 k^{2}

5 x - 19 \leq 1 or - 4 x + 3 < - 6

4 w^{2} + 28 w = 0