What is sqrt(x+1)-1<= x ?

The solution to sqrt(x+1)-1 = 0

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sqrt(x+1)-1<= x

x + 1 - 1 \leq x

x = - 1 or x \geq 0

Interval Notation

x = - 1 \cup [0, \infty)

Solution steps

x + 1 - 1 \leq x

Add

1

to both sides

x + 1 - 1 + 1 \leq x + 1

Simplify

x + 1 \leq x + 1

f (x) \leq g (x)

then

g (x) \geq 0

x + 1 \geq 0 : x \geq - 1

x + 1 \leq x + 1 : x \leq - 1 or x \geq 0

Find singularity points

Find non-negative values for radicals:

x \geq - 1

Combine the intervals

x \geq - 1 and (x \leq - 1 or x \geq 0) and x \geq - 1

Merge Overlapping Intervals

x = - 1 or x \geq 0

Verify Solutions:

x = - 1

True

, x \geq 0

True

The solution is

x = - 1 or x \geq 0

- 3 x - 7 < 14

5 x^{2} - 80 = 0

2 x^{2} - 2 x + \frac{1}{2} = 0

- 0.4 a + 3 = 7

5 x + 1 < - 4 or 6 x - 2 \geq 10