What is sqrt(4x-12)<= sqrt(2x+12) ?

The solution to sqrt(4x-12)<= sqrt(2x+12) is 3<= x<= 12

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sqrt(4x-12)<= sqrt(2x+12)

4 x - 12 \leq 2 x + 12

3 \leq x \leq 12

Interval Notation

[3, 12]

Solution steps

4 x - 12 \leq 2 x + 12

Square both sides

(4 x - 12)^{2} \leq (2 x + 12)^{2}

Simplify

4 x - 12 \leq 2 x + 12

4 x - 12 \leq 2 x + 12 : x \leq 12

x \leq 12

Find singularity points

Find non-negative values for radicals:

x \geq 3

Combine the intervals

x \leq 12 and x \geq 3

Merge Overlapping Intervals

3 \leq x \leq 12

7 ∣ 3 x ∣ + 2 = 16

- 8 (x - 4) - 3 \leq 5

(x - 4) (x + 1) < 0

\frac{x}{4} \geq - 1 and - 4 x - 4 \leq - 3

\frac{1}{4} x + x = - 3 + \frac{1}{2} x

What is sqrt(4x-12)<= sqrt(2x+12) ?
The solution to sqrt(4x-12)<= sqrt(2x+12) is 3<= x<= 12