What is sqrt(x-5)-sqrt(9-x)>1 ?

The solution to sqrt(x-5)-sqrt(9-x)>1 is (sqrt(7))/2+7<x<= 9

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sqrt(x-5)-sqrt(9-x)>1

x - 5 - 9 - x > 1

\frac{7}{2} + 7 < x \leq 9

Interval Notation

(\frac{7}{2} + 7, 9]

Decimal Notation

8.32287 \dots < x \leq 9

Solution steps

x - 5 - 9 - x > 1

Add

9 - x

to both sides

x - 5 - 9 - x + 9 - x > 1 + 9 - x

Simplify

x - 5 > 1 + 9 - x

f (x) > g (x)

then

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

For

1 + 9 - x < 0 :

No Solution

For

1 + 9 - x \geq 0 : x > \frac{7}{2} + 7

Find singularity points

Find non-negative values for radicals:

5 \leq x \leq 9

Combine the intervals

(x > \frac{7}{2} + 7 or No Solution) and 5 \leq x \leq 9

Merge Overlapping Intervals

\frac{7}{2} + 7 < x \leq 9

Verify Solutions:

\frac{7}{2} + 7 < x \leq 9

True

The solution is

\frac{7}{2} + 7 < x \leq 9

x^{2} < x + 1

x > 1 - x

x + 5 < x + 7

x^{2} - 6 x + 8 \leq x + 1

\frac{( x - 1 )}{( x - 2 )} < 0

What is sqrt(x-5)-sqrt(9-x)>1 ?
The solution to sqrt(x-5)-sqrt(9-x)>1 is (sqrt(7))/2+7<x<= 9