Question

solve the inequality and write the solution in interval notation. leave answers in exact, simplified form.

\\2x^2 + 4 \le 8x\\

Explanation:

Step1: Write in standard form

$$2x^2 - 8x + 4 \le 0$$

Step2: Simplify the inequality

$$x^2 - 4x + 2 \le 0$$

Step3: Find the boundary roots

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)} = 2 \pm \sqrt{2}$$

Step4: Determine the interval

$$[2 - \sqrt{2}, 2 + \sqrt{2}]$$

Answer:

\([2 - \sqrt{2}, 2 + \sqrt{2}]\)