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

The solution to sqrt(x(x+3))<= 6-x^2-3x is -4<= x<=-3\lor 0<= x<= 1

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

x (x + 3) \leq 6 - x^{2} - 3 x

- 4 \leq x \leq - 3 or 0 \leq x \leq 1

Interval Notation

[- 4, - 3] \cup [0, 1]

Solution steps

x (x + 3) \leq 6 - x^{2} - 3 x

f (x) \leq g (x)

then

g (x) \geq 0

6 - x^{2} - 3 x \geq 0 : \frac{- 33 - 3}{2} \leq x \leq \frac{33 - 3}{2}

x (x + 3) \leq 6 - x^{2} - 3 x : x \leq \frac{- 3 - 3 5}{2} or - 4 \leq x \leq 1 or x \geq \frac{3 5 - 3}{2}

Find singularity points

Find non-negative values for radicals:

x \leq - 3 or x \geq 0

Combine the intervals

\frac{- 33 - 3}{2} \leq x \leq \frac{33 - 3}{2} and (x \leq \frac{- 3 - 3 5}{2} or - 4 \leq x \leq 1 or x \geq \frac{3 5 - 3}{2}) and (x \leq - 3 or x \geq 0)

Merge Overlapping Intervals

- 4 \leq x \leq - 3 or 0 \leq x \leq 1

Verify Solutions:

- 4 \leq x \leq - 3

True

, 0 \leq x \leq 1

True

The solution is

- 4 \leq x \leq - 3 or 0 \leq x \leq 1

x^{2} - 10 x + 18 = 0

x^{2} = - 6 x

3 ∣ 5 + t ∣ - 1 \geq 26

(5 + i) + (1 - 3 i)

k^{2} - 13 k + 42

What is sqrt(x(x+3))<= 6-x^2-3x ?
The solution to sqrt(x(x+3))<= 6-x^2-3x is -4<= x<=-3\lor 0<= x<= 1