What is x-2sqrt(4-x)+1>= 0 ?

The solution to x-2sqrt(4-x)+1>= 0 is 2sqrt(6)-3<= x<= 4

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x-2sqrt(4-x)+1>= 0

x - 2 4 - x + 1 \geq 0

26 - 3 \leq x \leq 4

Interval Notation

[26 - 3, 4]

Decimal Notation

1.89897 \dots \leq x \leq 4

Solution steps

x - 2 4 - x + 1 \geq 0

Subtract

x + 1

from both sides

x - 2 4 - x + 1 - (x + 1) \geq 0 - (x + 1)

Simplify

- 2 4 - x \geq - (x + 1)

Multiply both sides by

- 1

2 4 - x \leq x + 1

Divide both sides by

2

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

f (x) \leq g (x)

then

g (x) \geq 0

\frac{x + 1}{2} \geq 0 : x \geq - 1

4 - x \leq \frac{x + 1}{2} : x \leq - 26 - 3 or x \geq 26 - 3

Find singularity points

Find non-negative values for radicals:

x \leq 4

Combine the intervals

x \geq - 1 and (x \leq - 26 - 3 or x \geq 26 - 3) and x \leq 4

Merge Overlapping Intervals

26 - 3 \leq x \leq 4

Verify Solutions:

26 - 3 \leq x \leq 4

True

The solution is

26 - 3 \leq x \leq 4

(a - 1)^{2} - (b - 2)^{2}

49 v^{2} - 1

\frac{4 ∣ x + 3 ∣}{3} = 8

\frac{2}{9} - [\frac{5}{4} + (\frac{7}{5} \cdot \frac{3}{2})] + \frac{2}{3}

x^{2} - 3 x = 2

What is x-2sqrt(4-x)+1>= 0 ?
The solution to x-2sqrt(4-x)+1>= 0 is 2sqrt(6)-3<= x<= 4