Solve (5x-1)<(3x^2)

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 13 + 5}{6}) \cup (\frac{13 + 5}{6}, + \infty)

Evaluate

5 x - 1 < 3 x^{2}

Move the expression to the left side

5 x - 1 - 3 x^{2} < 0

Rewrite the expression

5 x - 1 - 3 x^{2} = 0

Add or subtract both sides

5 x - 3 x^{2} = 1

Divide both sides

\frac{5 x - 3 x ^{2}}{- 3} = \frac{1}{- 3}

Evaluate

- \frac{5}{3} x + x^{2} = - \frac{1}{3}

Add the same value to both sides

- \frac{5}{3} x + x^{2} + \frac{25}{36} = - \frac{1}{3} + \frac{25}{36}

Simplify the expression

(x - \frac{5}{6})^{2} = \frac{13}{36}

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

x - \frac{5}{6} = \pm \frac{13}{36}

Simplify the expression

x - \frac{5}{6} = \pm \frac{13}{6}

Separate the equation into 2 possible cases

x - \frac{5}{6} = \frac{13}{6} x - \frac{5}{6} = - \frac{13}{6}

Solve the equation

More Steps

x = \frac{13 + 5}{6} x - \frac{5}{6} = - \frac{13}{6}

Solve the equation

More Steps

x = \frac{13 + 5}{6} x = \frac{- 13 + 5}{6}

Determine the test intervals using the critical values

x < \frac{- 13 + 5}{6} \frac{- 13 + 5}{6} < x < \frac{13 + 5}{6} x > \frac{13 + 5}{6}

Choose a value form each interval

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

To determine if x < \frac{- 13 + 5}{6} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < \frac{- 13 + 5}{6} is the solution x_{2} = 1 x_{3} = 2

To determine if \frac{- 13 + 5}{6} < x < \frac{13 + 5}{6} is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

More Steps

x < \frac{- 13 + 5}{6} is the solution \frac{- 13 + 5}{6} < x < \frac{13 + 5}{6} is not a solution x_{3} = 2

To determine if x > \frac{13 + 5}{6} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < \frac{- 13 + 5}{6} is the solution \frac{- 13 + 5}{6} < x < \frac{13 + 5}{6} is not a solution x > \frac{13 + 5}{6} is the solution

Solution

x \in (- \infty, \frac{- 13 + 5}{6}) \cup (\frac{13 + 5}{6}, + \infty)

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