What is 9-sqrt(6x+3)>= 6 ?

The solution to 9-sqrt(6x+3)>= 6 is -1/2 <= x<= 1

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9-sqrt(6x+3)>= 6

9 - 6 x + 3 \geq 6

- \frac{1}{2} \leq x \leq 1

Interval Notation

[- \frac{1}{2}, 1]

Decimal Notation

- 0.5 \leq x \leq 1

Solution steps

9 - 6 x + 3 \geq 6

Move

9

to the right side

- 6 x + 3 \geq - 3

Multiply both sides by

- 1

6 x + 3 \leq 3

Square both sides

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

Simplify

6 x + 3 \leq 9

6 x + 3 \leq 9 : x \leq 1

x \leq 1

Find singularity points

Find non-negative values for radicals:

x \geq - \frac{1}{2}

Combine the intervals

x \leq 1 and x \geq - \frac{1}{2}

Merge Overlapping Intervals

- \frac{1}{2} \leq x \leq 1

∣ 5 x - 15 ∣ \geq 20

\frac{1 0 ^{- 14}}{1.4 \cdot 1 0 ^{- 2}}

4 + 5 x - 2 x^{2} = 5

\frac{x - 5}{3} + \frac{x + 4}{2} \geq \frac{x + 3}{6}

(x^{4} y^{2} z) (y^{2} z)