What is x^3<= 6x^2+9x ?

The solution to x^3<= 6x^2+9x is x<= 3(1-sqrt(2))\lor 0<= x<= 3(1+sqrt(2))

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x^3<= 6x^2+9x

x^{3} \leq 6 x^{2} + 9 x

x \leq 3 (1 - 2) or 0 \leq x \leq 3 (1 + 2)

Interval Notation

(- \infty, 3 (1 - 2)] \cup [0, 3 (1 + 2)]

Decimal Notation

x \leq - 1.24264 \dots or 0 \leq x \leq 7.24264 \dots

Solution steps

x^{3} \leq 6 x^{2} + 9 x

Rewrite in standard form

x^{3} - 9 x - 6 x^{2} \leq 0

Factor

x^{3} - 9 x - 6 x^{2} : x (x - 3 (1 + 2)) (x - 3 (1 - 2))

x (x - 3 (1 + 2)) (x - 3 (1 - 2)) \leq 0

Identify the intervals

x \leq 3 (1 - 2) or 0 \leq x \leq 3 (1 + 2)

i^{84}

- 4 \leq x \leq 6

4 ∣ x - 4 ∣ > 12

- 2 (\frac{2}{3})^{2}

4 x^{2} - 20 x + 25 = 7

What is x^3<= 6x^2+9x ?
The solution to x^3<= 6x^2+9x is x<= 3(1-sqrt(2))\lor 0<= x<= 3(1+sqrt(2))