What is sqrt(3x-5)-sqrt(4-x)>1 ?

The solution to sqrt(3x-5)-sqrt(4-x)>1 is 3<x<= 4

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sqrt(3x-5)-sqrt(4-x)>1

3 x - 5 - 4 - x > 1

3 < x \leq 4

Interval Notation

(3, 4]

Solution steps

3 x - 5 - 4 - x > 1

Add

4 - x

to both sides

3 x - 5 - 4 - x + 4 - x > 1 + 4 - x

Simplify

3 x - 5 > 1 + 4 - x

f (x) > g (x)

then

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

For

1 + 4 - x < 0 :

No Solution

For

1 + 4 - x \geq 0 : x > 3

Find singularity points

Find non-negative values for radicals:

\frac{5}{3} \leq x \leq 4

Combine the intervals

(x > 3 or No Solution) and \frac{5}{3} \leq x \leq 4

Merge Overlapping Intervals

3 < x \leq 4

Verify Solutions:

3 < x \leq 4

True

The solution is

3 < x \leq 4

x^{2} = 21 + 4 x

x^{2} + 10 x y - 24 y^{2}

3 y^{2} - 19 y - 14

x^{2} - 14 x + 31 = 63

x^{2} - 18 x - 19

What is sqrt(3x-5)-sqrt(4-x)>1 ?
The solution to sqrt(3x-5)-sqrt(4-x)>1 is 3<x<= 4