The solution to 2x>= x^2 is 0<= x<= 2

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2x>= x^2

Solution

2 x \geq x^{2}

0 \leq x \leq 2

Interval Notation

[0, 2]

Solution steps

2 x \geq x^{2}

Rewrite in standard form

2 x - x^{2} \geq 0

Factor

2 x - x^{2} : - x (x - 2)

- x (x - 2) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x - 2)) (- 1) \leq 0 \cdot (- 1)

Simplify

x (x - 2) \leq 0

Identify the intervals

0 \leq x \leq 2

\frac{( x - 1 ) ( x - 2 ) ^{2}}{( x + 5 )} \geq 0

4 x^{2} - 9 \geq 0

x^{3} \geq 0

\frac{( x + 4 ) ( x - 10 )}{( x + 2 )} > 0

24 x^{6} - 10 x^{3} + 1 < 0