Solve (x-1)(x-4)>=2

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

x \in (- \infty, \frac{- 17 + 5}{2}] \cup [\frac{17 + 5}{2}, + \infty)

Evaluate

(x - 1) (x - 4) \geq 2

Move the expression to the left side

(x - 1) (x - 4) - 2 \geq 0

Subtract the terms

More Steps

x^{2} - 5 x + 2 \geq 0

Rewrite the expression

x^{2} - 5 x + 2 = 0

Add or subtract both sides

x^{2} - 5 x = - 2

Add the same value to both sides

x^{2} - 5 x + \frac{25}{4} = - 2 + \frac{25}{4}

Simplify the expression

(x - \frac{5}{2})^{2} = \frac{17}{4}

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

x - \frac{5}{2} = \pm \frac{17}{4}

Simplify the expression

x - \frac{5}{2} = \pm \frac{17}{2}

Separate the equation into 2 possible cases

x - \frac{5}{2} = \frac{17}{2} x - \frac{5}{2} = - \frac{17}{2}

Solve the equation

More Steps

x = \frac{17 + 5}{2} x - \frac{5}{2} = - \frac{17}{2}

Solve the equation

More Steps

x = \frac{17 + 5}{2} x = \frac{- 17 + 5}{2}

Determine the test intervals using the critical values

x < \frac{- 17 + 5}{2} \frac{- 17 + 5}{2} < x < \frac{17 + 5}{2} x > \frac{17 + 5}{2}

Choose a value form each interval

x_{1} = - 1 x_{2} = 3 x_{3} = 6

To determine if x < \frac{- 17 + 5}{2} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < \frac{- 17 + 5}{2} is the solution x_{2} = 3 x_{3} = 6

To determine if \frac{- 17 + 5}{2} < x < \frac{17 + 5}{2} is the solution to the inequality,test if the chosen value x = 3 satisfies the initial inequality

More Steps

x < \frac{- 17 + 5}{2} is the solution \frac{- 17 + 5}{2} < x < \frac{17 + 5}{2} is not a solution x_{3} = 6

To determine if x > \frac{17 + 5}{2} is the solution to the inequality,test if the chosen value x = 6 satisfies the initial inequality

More Steps

x < \frac{- 17 + 5}{2} is the solution \frac{- 17 + 5}{2} < x < \frac{17 + 5}{2} is not a solution x > \frac{17 + 5}{2} is the solution

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

x \leq \frac{- 17 + 5}{2} is the solution x \geq \frac{17 + 5}{2} is the solution

Solution

x \in (- \infty, \frac{- 17 + 5}{2}] \cup [\frac{17 + 5}{2}, + \infty)

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