Solve 2x^2 -3x ≥4

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 41 + 3}{4}] \cup [\frac{41 + 3}{4}, + \infty)

Evaluate

2 x^{2} - 3 x \geq 4

Move the expression to the left side

2 x^{2} - 3 x - 4 \geq 0

Rewrite the expression

2 x^{2} - 3 x - 4 = 0

Add or subtract both sides

2 x^{2} - 3 x = 4

Divide both sides

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

Evaluate

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

Add the same value to both sides

x^{2} - \frac{3}{2} x + \frac{9}{16} = 2 + \frac{9}{16}

Simplify the expression

(x - \frac{3}{4})^{2} = \frac{41}{16}

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

x - \frac{3}{4} = \pm \frac{41}{16}

Simplify the expression

x - \frac{3}{4} = \pm \frac{41}{4}

Separate the equation into 2 possible cases

x - \frac{3}{4} = \frac{41}{4} x - \frac{3}{4} = - \frac{41}{4}

Solve the equation

More Steps

x = \frac{41 + 3}{4} x - \frac{3}{4} = - \frac{41}{4}

Solve the equation

More Steps

x = \frac{41 + 3}{4} x = \frac{- 41 + 3}{4}

Determine the test intervals using the critical values

x < \frac{- 41 + 3}{4} \frac{- 41 + 3}{4} < x < \frac{41 + 3}{4} x > \frac{41 + 3}{4}

Choose a value form each interval

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

To determine if x < \frac{- 41 + 3}{4} is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < \frac{- 41 + 3}{4} is the solution x_{2} = 1 x_{3} = 3

To determine if \frac{- 41 + 3}{4} < x < \frac{41 + 3}{4} is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

More Steps

x < \frac{- 41 + 3}{4} is the solution \frac{- 41 + 3}{4} < x < \frac{41 + 3}{4} is not a solution x_{3} = 3

To determine if x > \frac{41 + 3}{4} is the solution to the inequality,test if the chosen value x = 3 satisfies the initial inequality

More Steps

x < \frac{- 41 + 3}{4} is the solution \frac{- 41 + 3}{4} < x < \frac{41 + 3}{4} is not a solution x > \frac{41 + 3}{4} is the solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

x \leq \frac{- 41 + 3}{4} is the solution x \geq \frac{41 + 3}{4} is the solution

Solution

x \in (- \infty, \frac{- 41 + 3}{4}] \cup [\frac{41 + 3}{4}, + \infty)

Show Solution