Solve 5 alpha /alpha^2 1 >=1

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

0 < α \leq 5

Alternative Form

α \in (0, 5]

Evaluate

(5 \times \frac{α}{α ^{2}}) \times 1 \geq 1

Find the domain

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(5 \times \frac{α}{α ^{2}}) \times 1 \geq 1, α \neq = 0

Remove the parentheses

5 \times \frac{α}{α ^{2}} \times 1 \geq 1

Simplify

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\frac{5}{α} \geq 1

Move the expression to the left side

\frac{5}{α} - 1 \geq 0

Subtract the terms

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\frac{5 - α}{α} \geq 0

Set the numerator and denominator of \frac{5 - α}{α} equal to 0 to find the values of α where sign changes may occur

5 - α = 0 α = 0

Calculate

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α = 5 α = 0

Determine the test intervals using the critical values

α < 0 0 < α < 5 α > 5

Choose a value form each interval

α_{1} = - 1 α_{2} = 3 α_{3} = 6

To determine if α < 0 is the solution to the inequality,test if the chosen value α = - 1 satisfies the initial inequality

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α < 0 is not a solution α_{2} = 3 α_{3} = 6

To determine if 0 < α < 5 is the solution to the inequality,test if the chosen value α = 3 satisfies the initial inequality

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α < 0 is not a solution 0 < α < 5 is the solution α_{3} = 6

To determine if α > 5 is the solution to the inequality,test if the chosen value α = 6 satisfies the initial inequality

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α < 0 is not a solution 0 < α < 5 is the solution α > 5 is not a solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

0 < α \leq 5 is the solution

The final solution of the original inequality is 0 < α \leq 5

0 < α \leq 5

Check if the solution is in the defined range

0 < α \leq 5, α \neq = 0

Solution

0 < α \leq 5

Alternative Form

α \in (0, 5]

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