What is the answer to |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6| ?

The answer to |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6| is x=1\lor 3<= x<= 4

Solve for x: |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6|

The solution is x=1\lor 3<= x<= 4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

The solution is x=1\lor 3<= x<= 4

x^{2} - 3 x + 2 - x^{2} - 5 x + 4 = 2 x^{2} - 8 x + 6

x = 1 or 3 \leq x \leq 4

Solution steps

x^{2} - 3 x + 2 - x^{2} - 5 x + 4 = 2 x^{2} - 8 x + 6

Find positive and negative intervals

x \leq 1, 1 < x < 2, 2 \leq x < 3, 3 \leq x < 4, x \geq 4

Solve the inequality for each interval

x = 1 or No Solution or 3 \leq x < 4 or x = 4

Merge Overlapping Intervals

x = 1 or 3 \leq x \leq 4

∣ 3 x - 1 ∣ = 4

∣ x ∣ + x = 0

∣ x - 1 ∣ = 4 - x

∣ x - 1 ∣ = ∣ x + 4 ∣

x^{2} + x + 2 - x^{2} - x + 1 = ∣ 2 x + 1 ∣

What is the answer to |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6| ?
The answer to |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6| is x=1\lor 3<= x<= 4
Solve for x: |x^2-3x+2|-|x^2-5x+4|=|2x^2-8x+6|
The solution is x=1\lor 3<= x<= 4