What is sqrt(2x+5)-sqrt(9+x)>0 ?

The solution to sqrt(2x+5)-sqrt(9+x)>0 is x>4

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sqrt(2x+5)-sqrt(9+x)>0

2 x + 5 - 9 + x > 0

x > 4

Interval Notation

(4, \infty)

Solution steps

2 x + 5 - 9 + x > 0

Add

9 + x

to both sides

2 x + 5 - 9 + x + 9 + x > 0 + 9 + x

Simplify

2 x + 5 > x + 9

f (x) > g (x)

then

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

For

x + 9 < 0 :

No Solution

For

x + 9 \geq 0 : x > 4

Find singularity points

Find non-negative values for radicals:

x \geq - \frac{5}{2}

Combine the intervals

(x > 4 or No Solution) and x \geq - \frac{5}{2}

Merge Overlapping Intervals

x > 4

\frac{a ^{2} - b ^{2}}{a - b}

(x + 3) 2 + 6 (x + 3) + 5 = 0

\frac{3 x - 3}{4} = \frac{2 x + 1}{3}

81 x^{2} - 9 y^{2}

2^{n} - 2^{n + 1}