Solve 9-4x>14x 9-12x^2-x

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 1801 + 43}{8}) \cup (\frac{1801 + 43}{8}, + \infty)

Evaluate

9 - 4 x > 14 x \times 9 - 12 x^{2} - x

Simplify

More Steps

9 - 4 x > 125 x - 12 x^{2}

Move the expression to the left side

9 - 4 x - (125 x - 12 x^{2}) > 0

Subtract the terms

More Steps

9 - 129 x + 12 x^{2} > 0

Rewrite the expression

9 - 129 x + 12 x^{2} = 0

Add or subtract both sides

- 129 x + 12 x^{2} = - 9

Divide both sides

\frac{- 129 x + 12 x ^{2}}{12} = \frac{- 9}{12}

Evaluate

- \frac{43}{4} x + x^{2} = - \frac{3}{4}

Add the same value to both sides

- \frac{43}{4} x + x^{2} + \frac{1849}{64} = - \frac{3}{4} + \frac{1849}{64}

Simplify the expression

(x - \frac{43}{8})^{2} = \frac{1801}{64}

Take the root of both sides of the equation and remember to use both positive and negative roots

x - \frac{43}{8} = \pm \frac{1801}{64}

Simplify the expression

x - \frac{43}{8} = \pm \frac{1801}{8}

Separate the equation into 2 possible cases

x - \frac{43}{8} = \frac{1801}{8} x - \frac{43}{8} = - \frac{1801}{8}

Solve the equation

More Steps

x = \frac{1801 + 43}{8} x - \frac{43}{8} = - \frac{1801}{8}

Solve the equation

More Steps

x = \frac{1801 + 43}{8} x = \frac{- 1801 + 43}{8}

Determine the test intervals using the critical values

x < \frac{- 1801 + 43}{8} \frac{- 1801 + 43}{8} < x < \frac{1801 + 43}{8} x > \frac{1801 + 43}{8}

Choose a value form each interval

x_{1} = - 1 x_{2} = 5 x_{3} = 12

To determine if x < \frac{- 1801 + 43}{8} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < \frac{- 1801 + 43}{8} is the solution x_{2} = 5 x_{3} = 12

To determine if \frac{- 1801 + 43}{8} < x < \frac{1801 + 43}{8} is the solution to the inequality,test if the chosen value x = 5 satisfies the initial inequality

More Steps

x < \frac{- 1801 + 43}{8} is the solution \frac{- 1801 + 43}{8} < x < \frac{1801 + 43}{8} is not a solution x_{3} = 12

To determine if x > \frac{1801 + 43}{8} is the solution to the inequality,test if the chosen value x = 12 satisfies the initial inequality

More Steps

x < \frac{- 1801 + 43}{8} is the solution \frac{- 1801 + 43}{8} < x < \frac{1801 + 43}{8} is not a solution x > \frac{1801 + 43}{8} is the solution

Solution

x \in (- \infty, \frac{- 1801 + 43}{8}) \cup (\frac{1801 + 43}{8}, + \infty)

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