Solve 2(p-4) 4(10p^2) < -10p-7p

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

p \in (- \infty, 0) \cup (\frac{- 1515 + 40}{20}, \frac{1515 + 40}{20})

Evaluate

2 (p - 4) \times 4 \times 10 p^{2} < - 10 p - 7 p

Multiply

More Steps

80 p^{2} (p - 4) < - 10 p - 7 p

Subtract the terms

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80 p^{2} (p - 4) < - 17 p

Move the expression to the left side

80 p^{2} (p - 4) - (- 17 p) < 0

Subtract the terms

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80 p^{3} - 320 p^{2} + 17 p < 0

Rewrite the expression

80 p^{3} - 320 p^{2} + 17 p = 0

Factor the expression

p (80 p^{2} - 320 p + 17) = 0

Separate the equation into 2 possible cases

p = 0 80 p^{2} - 320 p + 17 = 0

Solve the equation

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p = 0 p = \frac{1515 + 40}{20} p = \frac{- 1515 + 40}{20}

Determine the test intervals using the critical values

p < 0 0 < p < \frac{- 1515 + 40}{20} \frac{- 1515 + 40}{20} < p < \frac{1515 + 40}{20} p > \frac{1515 + 40}{20}

Choose a value form each interval

p_{1} = - 1 p_{2} = \frac{- 1515 + 40}{40} p_{3} = 2 p_{4} = 5

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

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p < 0 is the solution p_{2} = \frac{- 1515 + 40}{40} p_{3} = 2 p_{4} = 5

To determine if 0 < p < \frac{- 1515 + 40}{20} is the solution to the inequality,test if the chosen value p = \frac{- 1515 + 40}{40} satisfies the initial inequality

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p < 0 is the solution 0 < p < \frac{- 1515 + 40}{20} is not a solution p_{3} = 2 p_{4} = 5

To determine if \frac{- 1515 + 40}{20} < p < \frac{1515 + 40}{20} is the solution to the inequality,test if the chosen value p = 2 satisfies the initial inequality

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p < 0 is the solution 0 < p < \frac{- 1515 + 40}{20} is not a solution \frac{- 1515 + 40}{20} < p < \frac{1515 + 40}{20} is the solution p_{4} = 5

To determine if p > \frac{1515 + 40}{20} is the solution to the inequality,test if the chosen value p = 5 satisfies the initial inequality

More Steps

p < 0 is the solution 0 < p < \frac{- 1515 + 40}{20} is not a solution \frac{- 1515 + 40}{20} < p < \frac{1515 + 40}{20} is the solution p > \frac{1515 + 40}{20} is not a solution

Solution

p \in (- \infty, 0) \cup (\frac{- 1515 + 40}{20}, \frac{1515 + 40}{20})

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