What is sqrt(2x)<= sqrt(x^2) ?

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sqrt(2x)<= sqrt(x^2)

2 x \leq x^{2}

x = 0 or x \geq 2

Interval Notation

Solution steps

2 x \leq x^{2}

Square both sides

(2 x)^{2} \leq (x^{2})^{2}

Simplify

2 x \leq x^{2}

2 x \leq x^{2} : x \leq 0 or x \geq 2

x \leq 0 or x \geq 2

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

(x \leq 0 or x \geq 2) and x \geq 0

Merge Overlapping Intervals

x = 0 or x \geq 2

x - 4 < - 3

x^{2} - 4 x + 3 \geq 2 - x

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

2 x + 3 \geq 3 x + 4

x - 8 \geq 12 - x

What is sqrt(2x)<= sqrt(x^2) ?
The solution to sqrt(2x)<= sqrt(x^2) is x=0\lor x>= 2