Solve 6(x^2)-4x>2*2

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 7 + 1}{3}) \cup (\frac{7 + 1}{3}, + \infty)

Evaluate

6 x^{2} - 4 x > 2 \times 2

Multiply the numbers

6 x^{2} - 4 x > 4

Move the expression to the left side

6 x^{2} - 4 x - 4 > 0

Rewrite the expression

6 x^{2} - 4 x - 4 = 0

Add or subtract both sides

6 x^{2} - 4 x = 4

Divide both sides

\frac{6 x ^{2} - 4 x}{6} = \frac{4}{6}

Evaluate

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

Add the same value to both sides

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

Simplify the expression

(x - \frac{1}{3})^{2} = \frac{7}{9}

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

x - \frac{1}{3} = \pm \frac{7}{9}

Simplify the expression

x - \frac{1}{3} = \pm \frac{7}{3}

Separate the equation into 2 possible cases

x - \frac{1}{3} = \frac{7}{3} x - \frac{1}{3} = - \frac{7}{3}

Solve the equation

More Steps

x = \frac{7 + 1}{3} x - \frac{1}{3} = - \frac{7}{3}

Solve the equation

More Steps

x = \frac{7 + 1}{3} x = \frac{- 7 + 1}{3}

Determine the test intervals using the critical values

x < \frac{- 7 + 1}{3} \frac{- 7 + 1}{3} < x < \frac{7 + 1}{3} x > \frac{7 + 1}{3}

Choose a value form each interval

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

To determine if x < \frac{- 7 + 1}{3} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < \frac{- 7 + 1}{3} is the solution x_{2} = 1 x_{3} = 2

To determine if \frac{- 7 + 1}{3} < x < \frac{7 + 1}{3} is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

More Steps

x < \frac{- 7 + 1}{3} is the solution \frac{- 7 + 1}{3} < x < \frac{7 + 1}{3} is not a solution x_{3} = 2

To determine if x > \frac{7 + 1}{3} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < \frac{- 7 + 1}{3} is the solution \frac{- 7 + 1}{3} < x < \frac{7 + 1}{3} is not a solution x > \frac{7 + 1}{3} is the solution

Solution

x \in (- \infty, \frac{- 7 + 1}{3}) \cup (\frac{7 + 1}{3}, + \infty)

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