What is -x-sqrt(x^2-8x+12)<0 ?

The solution to -x-sqrt(x^2-8x+12) = 6

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-x-sqrt(x^2-8x+12)<0

- x - x^{2} - 8 x + 12 < 0

x \leq 2 or x \geq 6

Interval Notation

(- \infty, 2] \cup [6, \infty)

Solution steps

- x - x^{2} - 8 x + 12 < 0

Add

x

to both sides

- x - x^{2} - 8 x + 12 + x < 0 + x

Simplify

- x^{2} - 8 x + 12 < x

Multiply both sides by -1 (reverse the inequality)

(- x^{2} - 8 x + 12) (- 1) > x (- 1)

Simplify

x^{2} - 8 x + 12 > - x

f (x) > g (x)

then

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

For

- x < 0 : x > 0

For

- x \geq 0 : x \leq 0

Find singularity points

Find non-negative values for radicals:

x \leq 2 or x \geq 6

Combine the intervals

(x \leq 0 or x > 0) and (x \leq 2 or x \geq 6)

Merge Overlapping Intervals

x \leq 2 or x \geq 6

Verify Solutions:

x \leq 2

True

, x \geq 6

True

The solution is

x \leq 2 or x \geq 6

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

y^{2} + 10 y = - 9

2 x^{2} - 5 x = 1

2 u - 10 = 8

\frac{3 + \frac{a}{b}}{6 - \frac{a}{b}}

What is -x-sqrt(x^2-8x+12)<0 ?
The solution to -x-sqrt(x^2-8x+12)<0 is x<= 2\lor x>= 6