What is |x^2-4x|>= 4 ?

The solution to |x^2-4x|>= 4 is x = 2sqrt(2)+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

|x^2-4x|>= 4

x^{2} - 4 x \geq 4

x \leq - 22 + 2 or x = 2 or x \geq 22 + 2

Interval Notation

(- \infty, - 22 + 2] \cup x = 2 \cup [22 + 2, \infty)

Decimal Notation

x \leq - 0.82842 \dots or x = 2 or x \geq 4.82842 \dots

Solution steps

x^{2} - 4 x \geq 4

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

x^{2} - 4 x \leq - 4 or x^{2} - 4 x \geq 4

x^{2} - 4 x \leq - 4 or x^{2} - 4 x \geq 4

x = 2 or (x \leq - 22 + 2 or x \geq 22 + 2)

Merge Overlapping Intervals

x \leq - 22 + 2 or x = 2 or x \geq 22 + 2

- 8 \cdot 4

3 x^{2} + 7 x + 4

[(\frac{3}{5})^{0}]^{- 2}

(- \frac{4}{5})^{2} + y^{2} = 1

- x^{3} + 4 x^{2} + 15 x - 18

What is |x^2-4x|>= 4 ?
The solution to |x^2-4x|>= 4 is x<=-2sqrt(2)+2\lor x=2\lor x>= 2sqrt(2)+2