What is sqrt(4-x)>= sqrt(3x+4) ?

The solution to sqrt(4-x)>= sqrt(3x+4) is -4/3 <= x<= 0

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sqrt(4-x)>= sqrt(3x+4)

4 - x \geq 3 x + 4

- \frac{4}{3} \leq x \leq 0

Interval Notation

[- \frac{4}{3}, 0]

Decimal Notation

- 1.33333 \dots \leq x \leq 0

Solution steps

4 - x \geq 3 x + 4

Square both sides

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

Simplify

4 - x \geq 3 x + 4

4 - x \geq 3 x + 4 : x \leq 0

x \leq 0

Find singularity points

Find non-negative values for radicals:

- \frac{4}{3} \leq x \leq 4

Combine the intervals

x \leq 0 and - \frac{4}{3} \leq x \leq 4

Merge Overlapping Intervals

- \frac{4}{3} \leq x \leq 0

- 8 \geq \frac{u}{2}

9 x^{2} - 6 y

k^{2} - 4

72 = 12 (2 y + 2)

(5 + 2 i) (5 - 2 i)

What is sqrt(4-x)>= sqrt(3x+4) ?
The solution to sqrt(4-x)>= sqrt(3x+4) is -4/3 <= x<= 0