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

The solution to sqrt(x^2+|x|-12)<x-2 is 3<= x< 16/5

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sqrt(x^2+|x|-12)<x-2

x^{2} + ∣ x ∣ - 12 < x - 2

3 \leq x < \frac{16}{5}

Interval Notation

[3, \frac{16}{5})

Decimal Notation

3 \leq x < 3.2

Solution steps

x^{2} + ∣ x ∣ - 12 < x - 2

f (x) < g (x)

then

g (x) > 0

x - 2 > 0 : x > 2

x^{2} + ∣ x ∣ - 12 < x - 2 : x < \frac{16}{5}

Find singularity points

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Combine the intervals

x > 2 and x < \frac{16}{5} and (x \leq - 3 or x \geq 3)

Merge Overlapping Intervals

3 \leq x < \frac{16}{5}

Verify Solutions:

3 \leq x < \frac{16}{5}

True

The solution is

3 \leq x < \frac{16}{5}

4 (2 y + 5) = 3 (5 y - 2)

\frac{6 y ^{3}}{4 y} \cdot \frac{16}{2 x ^{2}}

0. 5^{- 1}

(- 3 d^{2} f^{3} g)^{2} [(- 3 d^{2} f)^{3}]^{2}

88.1 = b + 3.236 (0)

What is sqrt(x^2+|x|-12)<x-2 ?
The solution to sqrt(x^2+|x|-12)<x-2 is 3<= x< 16/5