What is -x-sqrt(x^2-4)<1 ?

The solution to -x-sqrt(x^2-4) = 2

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-x-sqrt(x^2-4)<1

- x - x^{2} - 4 < 1

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

Interval Notation

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

Decimal Notation

x < - 2.5 or x \geq 2

Solution steps

- x - x^{2} - 4 < 1

Add

x

to both sides

- x - x^{2} - 4 + x < 1 + x

Simplify

- x^{2} - 4 < 1 + x

Multiply both sides by -1 (reverse the inequality)

(- x^{2} - 4) (- 1) > 1 \cdot (- 1) + x (- 1)

Simplify

x^{2} - 4 > - 1 - x

f (x) > g (x)

then

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

For

- 1 - x < 0 : x > - 1

For

- 1 - x \geq 0 : x < - \frac{5}{2}

Find singularity points

Find non-negative values for radicals:

x \leq - 2 or x \geq 2

Combine the intervals

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

Merge Overlapping Intervals

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

Verify Solutions:

x < - \frac{5}{2}

True

, x \geq 2

True

The solution is

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

a^{2} + 1 4^{2} = 2 8^{2}

3 a^{2} - 36 = 81

7 x + 1 = 9 x - 7

\frac{h ^{4}}{h ^{6}}

2 x^{2} - 13 x + 6 = 0

What is -x-sqrt(x^2-4)<1 ?
The solution to -x-sqrt(x^2-4)<1 is x<-5/2 \lor x>= 2