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

The solution to sqrt(x)>= 4-x is x>= (-sqrt(17)+9)/2

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sqrt(x)>= 4-x

x \geq 4 - x

x \geq \frac{- 17 + 9}{2}

Interval Notation

[\frac{- 17 + 9}{2}, \infty)

Decimal Notation

x \geq 2.43844 \dots

Solution steps

x \geq 4 - x

f (x) \geq g (x)

then

g (x) < 0 or g (x) \geq 0

For

4 - x < 0 : x > 4

For

4 - x \geq 0 : \frac{- 17 + 9}{2} \leq x \leq 4

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

(\frac{- 17 + 9}{2} \leq x \leq 4 or x > 4) and x \geq 0

Merge Overlapping Intervals

x \geq \frac{- 17 + 9}{2}

Verify Solutions:

x \geq \frac{- 17 + 9}{2}

True

The solution is

x \geq \frac{- 17 + 9}{2}

(3) x + 3 + x^{2} + 4 x - 5 > 0

x + 2 < x + 5

2 - x > x

\frac{( 2 x - 1 )}{( x - 2 )} < 1

1 - x^{2} + 1 < 3 - x^{2}