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

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

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

x^{2} + x - 12 < x

3 \leq x < 12

Interval Notation

[3, 12)

Solution steps

x^{2} + x - 12 < x

f (x) < g (x)

then

g (x) > 0

x > 0

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

Find singularity points

Find non-negative values for radicals:

x \leq - 4 or x \geq 3

Combine the intervals

x > 0 and x < 12 and (x \leq - 4 or x \geq 3)

Merge Overlapping Intervals

3 \leq x < 12

Verify Solutions:

3 \leq x < 12

True

The solution is

3 \leq x < 12

5 - 2 x < 6 x - 1

6 x - x^{2} < 3 + x

\frac{- 3}{( x ^{2} + 2 x + 10 )} \geq 0

\frac{( x + 5 )}{∣ 1 - x ∣} > 1

5 - x < 3