What is (m+2)+(4m+5)^{1/2}>1 ?

The solution to (m+2)+(4m+5)^{1/2}>1 is m>-5^{1/2}+1

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(m+2)+(4m+5)^{1/2}>1

(m + 2) + (4 m + 5)^{\frac{1}{2}} > 1

m > - 5^{\frac{1}{2}} + 1

Interval Notation

(- 5^{\frac{1}{2}} + 1, \infty)

Decimal Notation

m > - 1.23606 \dots

Solution steps

m + 2 + (4 m + 5)^{\frac{1}{2}} > 1

Simplify

m + 2 + (4 m + 5)^{\frac{1}{2}} > 1 : m + 2 + 4 m + 5 > 1

m + 2 + 4 m + 5 > 1

Subtract

m + 2

from both sides

m + 2 + 4 m + 5 - (m + 2) > 1 - (m + 2)

Simplify

4 m + 5 > - m - 1

f (x) > g (x)

then

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

For

- m - 1 < 0 : m > - 1

For

- m - 1 \geq 0 : - 5^{\frac{1}{2}} + 1 < m \leq - 1

Find singularity points

Find non-negative values for radicals:

m \geq - \frac{5}{4}

Combine the intervals

(- 5^{\frac{1}{2}} + 1 < m \leq - 1 or m > - 1) and m \geq - \frac{5}{4}

Merge Overlapping Intervals

m > - 5^{\frac{1}{2}} + 1

Verify Solutions:

m > - 5^{\frac{1}{2}} + 1

True

The solution is

m > - 5^{\frac{1}{2}} + 1

x + 3 < x + 33

x < 2

\frac{x}{2} > 4

x < 2 - x

- x^{2} + 8 x - 7 < x - 1

What is (m+2)+(4m+5)^{1/2}>1 ?
The solution to (m+2)+(4m+5)^{1/2}>1 is m>-5^{1/2}+1