Solve (1-a^2)(2a-a^2)>0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

a \in (- \infty, - 1) \cup (0, 1) \cup (2, + \infty)

Evaluate

(1 - a^{2}) (2 a - a^{2}) > 0

Rewrite the expression

(1 - a^{2}) (2 a - a^{2}) = 0

Separate the equation into 2 possible cases

1 - a^{2} = 0 2 a - a^{2} = 0

Solve the equation

More Steps

a = 1 a = - 1 2 a - a^{2} = 0

Solve the equation

More Steps

a = 1 a = - 1 a = 0 a = 2

Determine the test intervals using the critical values

a < - 1 - 1 < a < 0 0 < a < 1 1 < a < 2 a > 2

Choose a value form each interval

a_{1} = - 2 a_{2} = - \frac{1}{2} a_{3} = \frac{1}{2} a_{4} = \frac{3}{2} a_{5} = 3

To determine if a < - 1 is the solution to the inequality,test if the chosen value a = - 2 satisfies the initial inequality

More Steps

a < - 1 is the solution a_{2} = - \frac{1}{2} a_{3} = \frac{1}{2} a_{4} = \frac{3}{2} a_{5} = 3

To determine if - 1 < a < 0 is the solution to the inequality,test if the chosen value a = - \frac{1}{2} satisfies the initial inequality

More Steps

a < - 1 is the solution - 1 < a < 0 is not a solution a_{3} = \frac{1}{2} a_{4} = \frac{3}{2} a_{5} = 3

To determine if 0 < a < 1 is the solution to the inequality,test if the chosen value a = \frac{1}{2} satisfies the initial inequality

More Steps

a < - 1 is the solution - 1 < a < 0 is not a solution 0 < a < 1 is the solution a_{4} = \frac{3}{2} a_{5} = 3

To determine if 1 < a < 2 is the solution to the inequality,test if the chosen value a = \frac{3}{2} satisfies the initial inequality

More Steps

a < - 1 is the solution - 1 < a < 0 is not a solution 0 < a < 1 is the solution 1 < a < 2 is not a solution a_{5} = 3

To determine if a > 2 is the solution to the inequality,test if the chosen value a = 3 satisfies the initial inequality

More Steps

a < - 1 is the solution - 1 < a < 0 is not a solution 0 < a < 1 is the solution 1 < a < 2 is not a solution a > 2 is the solution

Solution

a \in (- \infty, - 1) \cup (0, 1) \cup (2, + \infty)

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