What is sqrt(2x+5)>x+2 ?

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

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sqrt(2x+5)>x+2

2 x + 5 > x + 2

- \frac{5}{2} \leq x < 2 - 1

Interval Notation

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

Decimal Notation

- 2.5 \leq x < 0.41421 \dots

Solution steps

2 x + 5 > x + 2

f (x) > g (x)

then

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

For

x + 2 < 0 : x < - 2

For

x + 2 \geq 0 : - 2 \leq x < 2 - 1

Find singularity points

Find non-negative values for radicals:

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

Combine the intervals

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

Merge Overlapping Intervals

- \frac{5}{2} \leq x < 2 - 1

Verify Solutions:

- \frac{5}{2} \leq x < 2 - 1

True

The solution is

- \frac{5}{2} \leq x < 2 - 1

x^{2} + 4 x - 12 > 0

(x^{\frac{3}{2}})^{2}

\frac{- 18}{9}

125 + x^{3}

28 x^{3} - 84 x^{2} + 63 x