What is sqrt(x+2)<= x+1 ?

The solution to sqrt(x+2) = (sqrt(5)-1)/2

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sqrt(x+2)<= x+1

x + 2 \leq x + 1

x \geq \frac{5 - 1}{2}

Interval Notation

[\frac{5 - 1}{2}, \infty)

Decimal Notation

x \geq 0.61803 \dots

Solution steps

x + 2 \leq x + 1

f (x) \leq g (x)

then

g (x) \geq 0

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

x + 2 \leq x + 1 : x \leq \frac{- 5 - 1}{2} or x \geq \frac{5 - 1}{2}

Find singularity points

Find non-negative values for radicals:

x \geq - 2

Combine the intervals

x \geq - 1 and (x \leq \frac{- 5 - 1}{2} or x \geq \frac{5 - 1}{2}) and x \geq - 2

Merge Overlapping Intervals

x \geq \frac{5 - 1}{2}

Verify Solutions:

x \geq \frac{5 - 1}{2}

True

The solution is

x \geq \frac{5 - 1}{2}

ln (x) = 5

(x + 4) (x - 5) \geq 0

2 x^{2} - 5 x < 12

2 x < 3 x + 4 \leq x + 6

\frac{( x ^{4} + x ^{3} + 8 x ^{2} - x + 1 )}{( x ^{2} - 1 )}