What is sqrt(x-1)-sqrt(8-2x)>= 1 ?

The solution to sqrt(x-1)-sqrt(8-2x)>= 1 is 34/9 <= x<= 4

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sqrt(x-1)-sqrt(8-2x)>= 1

x - 1 - 8 - 2 x \geq 1

\frac{34}{9} \leq x \leq 4

Interval Notation

[\frac{34}{9}, 4]

Decimal Notation

3.77777 \dots \leq x \leq 4

Solution steps

x - 1 - 8 - 2 x \geq 1

Add

8 - 2 x

to both sides

x - 1 - 8 - 2 x + 8 - 2 x \geq 1 + 8 - 2 x

Simplify

x - 1 \geq 1 + 8 - 2 x

f (x) \geq g (x)

then

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

For

1 + 8 - 2 x < 0 :

No Solution

For

1 + 8 - 2 x \geq 0 : x \geq \frac{34}{9}

Find singularity points

Find non-negative values for radicals:

1 \leq x \leq 4

Combine the intervals

(x \geq \frac{34}{9} or No Solution) and 1 \leq x \leq 4

Merge Overlapping Intervals

\frac{34}{9} \leq x \leq 4

Verify Solutions:

\frac{34}{9} \leq x \leq 4

True

The solution is

\frac{34}{9} \leq x \leq 4

\frac{2}{( 3 x - 5 )} \geq 0

3 - x > 3 (1 - x^{2})^{\frac{1}{2}}

x + 5 - 25 - 10 x + x^{2} \leq 2 x

2 x \leq x^{2}

x - 4 < - 3

What is sqrt(x-1)-sqrt(8-2x)>= 1 ?
The solution to sqrt(x-1)-sqrt(8-2x)>= 1 is 34/9 <= x<= 4