What is sqrt(3x)<= sqrt(2x+3) ?

The solution to sqrt(3x)<= sqrt(2x+3) is 0<= x<= 3

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sqrt(3x)<= sqrt(2x+3)

3 x \leq 2 x + 3

0 \leq x \leq 3

Interval Notation

[0, 3]

Solution steps

3 x \leq 2 x + 3

Square both sides

(3 x)^{2} \leq (2 x + 3)^{2}

Simplify

3 x \leq 2 x + 3

Move

2 x

to the left side

x \leq 3

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x \leq 3 and x \geq 0

Merge Overlapping Intervals

0 \leq x \leq 3

- 6 \leq - 2 x + 2 < 10

2 c^{2} - 7 c + 5

2 x^{2} - 15 x - 1 = 0

\frac{x + 8}{3} = \frac{3 x - 2}{8}

(2 t - 5) (5 t - 2) = 0

What is sqrt(3x)<= sqrt(2x+3) ?
The solution to sqrt(3x)<= sqrt(2x+3) is 0<= x<= 3