What is 3+sqrt(3x+12)<= 9 ?

The solution to 3+sqrt(3x+12)<= 9 is -4<= x<= 8

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3+sqrt(3x+12)<= 9

3 + 3 x + 12 \leq 9

- 4 \leq x \leq 8

Interval Notation

[- 4, 8]

Solution steps

3 + 3 x + 12 \leq 9

Move

3

to the right side

3 x + 12 \leq 6

Square both sides

(3 x + 12)^{2} \leq 6^{2}

Simplify

3 x + 12 \leq 36

3 x + 12 \leq 36 : x \leq 8

x \leq 8

Find singularity points

Find non-negative values for radicals:

x \geq - 4

Combine the intervals

x \leq 8 and x \geq - 4

Merge Overlapping Intervals

- 4 \leq x \leq 8

3 x^{2} - 27 = 0

2 x^{2} - 15 x + 27

5 - ∣ x ∣ = 15

24 a^{2} b^{5} c^{8} + 30 a^{9} b^{3} - 22 a^{7} b c^{4}

- 14 > 7 v