What is sqrt(6x-x^2)<3+x ?

The solution to sqrt(6x-x^2)<3+x is 0<= x<= 6

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sqrt(6x-x^2)<3+x

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

0 \leq x \leq 6

Interval Notation

[0, 6]

Solution steps

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

f (x) < g (x)

then

g (x) > 0

3 + x > 0 : x > - 3

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

True for all

x

Find singularity points

Find non-negative values for radicals:

0 \leq x \leq 6

Combine the intervals

x > - 3 and True for all x and 0 \leq x \leq 6

Merge Overlapping Intervals

0 \leq x \leq 6

Verify Solutions:

0 \leq x \leq 6

True

The solution is

0 \leq x \leq 6

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

\frac{( x + 5 )}{∣ 1 - x ∣} > 1

5 - x < 3

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

x - 1 < 7 - x