What is |((x^2-5x+4))/((x^2-4))|>= 1 ?

The solution to |((x^2-5x+4))/((x^2-4))|>= 1 is x<-2\lor-2<x<= 0\lor 8/5 <= x<2\lor 2<x<= 5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

|((x^2-5x+4))/((x^2-4))|>= 1

\frac{( x ^{2} - 5 x + 4 )}{( x ^{2} - 4 )} \geq 1

x < - 2 or - 2 < x \leq 0 or \frac{8}{5} \leq x < 2 or 2 < x \leq \frac{5}{2}

Interval Notation

(- \infty, - 2) \cup (- 2, 0] \cup [\frac{8}{5}, 2) \cup (2, \frac{5}{2}]

Decimal Notation

x < - 2 or - 2 < x \leq 0 or 1.6 \leq x < 2 or 2 < x \leq 2.5

Solution steps

\frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \geq 1

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

\frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \leq - 1 or \frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \geq 1

\frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \leq - 1 or \frac{x ^{2} - 5 x + 4}{x ^{2} - 4} \geq 1

(- 2 < x \leq 0 or 2 < x \leq \frac{5}{2}) or (x < - 2 or \frac{8}{5} \leq x < 2)

Merge Overlapping Intervals

x < - 2 or - 2 < x \leq 0 or \frac{8}{5} \leq x < 2 or 2 < x \leq \frac{5}{2}

\frac{( x ^{2} + 12 )}{( 7 x )} \geq 1

\frac{( ∣ x ∣ - 4 )}{( ∣ x ∣ - 5 )} \leq 0

∣ x ∣ - ∣ x - 2 ∣ \geq 1

\frac{3 - x}{2} \leq 2

\frac{( 2 x - 4 )}{- 6} + 2 \leq 10

What is |((x^2-5x+4))/((x^2-4))|>= 1 ?
The solution to |((x^2-5x+4))/((x^2-4))|>= 1 is x<-2\lor-2<x<= 0\lor 8/5 <= x<2\lor 2<x<= 5/2