What is (x-8)^{1/2}>= 12-x ?

The solution to (x-8)^{1/2}>= 12-x is x>= (-17^{1/2}+25)/2

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(x-8)^{1/2}>= 12-x

(x - 8)^{\frac{1}{2}} \geq 12 - x

x \geq \frac{- 1 7 ^{\frac{1}{2}} + 25}{2}

Interval Notation

[\frac{- 1 7 ^{\frac{1}{2}} + 25}{2}, \infty)

Decimal Notation

x \geq 10.43844 \dots

Solution steps

(x - 8)^{\frac{1}{2}} \geq 12 - x

Simplify

(x - 8)^{\frac{1}{2}} \geq 12 - x : x - 8 \geq 12 - x

x - 8 \geq 12 - x

f (x) \geq g (x)

then

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

For

12 - x < 0 : x > 12

For

12 - x \geq 0 : \frac{- 1 7 ^{\frac{1}{2}} + 25}{2} \leq x \leq 12

Find singularity points

Find non-negative values for radicals:

x \geq 8

Combine the intervals

(\frac{- 1 7 ^{\frac{1}{2}} + 25}{2} \leq x \leq 12 or x > 12) and x \geq 8

Merge Overlapping Intervals

x \geq \frac{- 1 7 ^{\frac{1}{2}} + 25}{2}

Verify Solutions:

x \geq \frac{- 1 7 ^{\frac{1}{2}} + 25}{2}

True

The solution is

x \geq \frac{- 1 7 ^{\frac{1}{2}} + 25}{2}

2 x < x^{2}

582 x^{2} - 9 x^{2}, 5 - 3 x + y + 6

2 x^{2} - 9 x^{2}, 5 - 3 x + y + 6

582 x^{2} - 9 x^{2}, 5 - 3 x + v + 6

2 x^{2} - 9 x^{2}, 5 - 3 x + v + 6

What is (x-8)^{1/2}>= 12-x ?
The solution to (x-8)^{1/2}>= 12-x is x>= (-17^{1/2}+25)/2