What is sqrt(x^2+x)>1-2x ?

The solution to sqrt(x^2+x)>1-2x is x>(-sqrt(13)+5)/6

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sqrt(x^2+x)>1-2x

x^{2} + x > 1 - 2 x

x > \frac{- 13 + 5}{6}

Interval Notation

(\frac{- 13 + 5}{6}, \infty)

Decimal Notation

x > 0.23240 \dots

Solution steps

x^{2} + x > 1 - 2 x

f (x) > g (x)

then

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

For

1 - 2 x < 0 : x > \frac{1}{2}

For

1 - 2 x \geq 0 : \frac{- 13 + 5}{6} < x \leq \frac{1}{2}

Find singularity points

Find non-negative values for radicals:

x \leq - 1 or x \geq 0

Combine the intervals

(\frac{- 13 + 5}{6} < x \leq \frac{1}{2} or x > \frac{1}{2}) and (x \leq - 1 or x \geq 0)

Merge Overlapping Intervals

x > \frac{- 13 + 5}{6}

Verify Solutions:

x > \frac{- 13 + 5}{6}

True

The solution is

x > \frac{- 13 + 5}{6}

x - 1 < 7 - x

n - 3 \geq 3 n + 1

x + 6 < x - 6

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

x + 3 < x + 33

What is sqrt(x^2+x)>1-2x ?
The solution to sqrt(x^2+x)>1-2x is x>(-sqrt(13)+5)/6