What is sqrt(x^2-3x-10)>x-2 ?

The solution to sqrt(x^2-3x-10)>x-2 is x 14

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sqrt(x^2-3x-10)>x-2

x^{2} - 3 x - 10 > x - 2

x \leq - 2 or x > 14

Interval Notation

(- \infty, - 2] \cup (14, \infty)

Solution steps

x^{2} - 3 x - 10 > x - 2

f (x) > g (x)

then

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

For

x - 2 < 0 : x < 2

For

x - 2 \geq 0 : x > 14

Find singularity points

Find non-negative values for radicals:

x \leq - 2 or x \geq 5

Combine the intervals

(x > 14 or x < 2) and (x \leq - 2 or x \geq 5)

Merge Overlapping Intervals

x \leq - 2 or x > 14

Verify Solutions:

x \leq - 2

True

, x > 14

True

The solution is

x \leq - 2 or x > 14

2 x^{3} + 14 x^{2} - 72

\frac{4}{1 - i} + \frac{2 - i}{1 + i}

0.5 x^{2} + 1 \leq 4 ∣ x ∣

2 x \cdot x^{3}

2 x^{2} + x - 1

What is sqrt(x^2-3x-10)>x-2 ?
The solution to sqrt(x^2-3x-10)>x-2 is x<=-2\lor x>14