What is sqrt(x^2+5x+2)>-1 ?

The solution to sqrt(x^2+5x+2)>-1 is x = (sqrt(17)-5)/2

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sqrt(x^2+5x+2)>-1

x^{2} + 5 x + 2 > - 1

x \leq \frac{- 17 - 5}{2} or x \geq \frac{17 - 5}{2}

Interval Notation

(- \infty, \frac{- 17 - 5}{2}] \cup [\frac{17 - 5}{2}, \infty)

Decimal Notation

x \leq - 4.56155 \dots or x \geq - 0.43844 \dots

Solution steps

x^{2} + 5 x + 2 > - 1

If n is even,

for all

u

True for all x

Find singularity points

Find non-negative values for radicals:

x \leq \frac{- 17 - 5}{2} or x \geq \frac{17 - 5}{2}

Combine the intervals

True for all x and (x \leq \frac{- 17 - 5}{2} or x \geq \frac{17 - 5}{2})

Merge Overlapping Intervals

x \leq \frac{- 17 - 5}{2} or x \geq \frac{17 - 5}{2}

25 = 1 x - 0.3 x

4 x^{2} = 27 x + 81

- 2 = - 4

5^{x} \cdot 5^{x}

(10 - 22 i) + (22 i)

What is sqrt(x^2+5x+2)>-1 ?
The solution to sqrt(x^2+5x+2)>-1 is x<= (-sqrt(17)-5)/2 \lor x>= (sqrt(17)-5)/2