Solve 4x^2 -12x-1>0

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 10 + 3}{2}) \cup (\frac{10 + 3}{2}, + \infty)

Evaluate

4 x^{2} - 12 x - 1 > 0

Rewrite the expression

4 x^{2} - 12 x - 1 = 0

Add or subtract both sides

4 x^{2} - 12 x = 1

Divide both sides

\frac{4 x ^{2} - 12 x}{4} = \frac{1}{4}

Evaluate

x^{2} - 3 x = \frac{1}{4}

Add the same value to both sides

x^{2} - 3 x + \frac{9}{4} = \frac{1}{4} + \frac{9}{4}

Simplify the expression

(x - \frac{3}{2})^{2} = \frac{5}{2}

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

x - \frac{3}{2} = \pm \frac{5}{2}

Simplify the expression

x - \frac{3}{2} = \pm \frac{10}{2}

Separate the equation into 2 possible cases

x - \frac{3}{2} = \frac{10}{2} x - \frac{3}{2} = - \frac{10}{2}

Solve the equation

More Steps

x = \frac{10 + 3}{2} x - \frac{3}{2} = - \frac{10}{2}

Solve the equation

More Steps

x = \frac{10 + 3}{2} x = \frac{- 10 + 3}{2}

Determine the test intervals using the critical values

x < \frac{- 10 + 3}{2} \frac{- 10 + 3}{2} < x < \frac{10 + 3}{2} x > \frac{10 + 3}{2}

Choose a value form each interval

x_{1} = - 1 x_{2} = 2 x_{3} = 4

To determine if x < \frac{- 10 + 3}{2} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < \frac{- 10 + 3}{2} is the solution x_{2} = 2 x_{3} = 4

To determine if \frac{- 10 + 3}{2} < x < \frac{10 + 3}{2} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < \frac{- 10 + 3}{2} is the solution \frac{- 10 + 3}{2} < x < \frac{10 + 3}{2} is not a solution x_{3} = 4

To determine if x > \frac{10 + 3}{2} is the solution to the inequality,test if the chosen value x = 4 satisfies the initial inequality

More Steps

x < \frac{- 10 + 3}{2} is the solution \frac{- 10 + 3}{2} < x < \frac{10 + 3}{2} is not a solution x > \frac{10 + 3}{2} is the solution

Solution

x \in (- \infty, \frac{- 10 + 3}{2}) \cup (\frac{10 + 3}{2}, + \infty)

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