What is the answer to |x^2-1|+|2-x^2|=1 ?

The answer to |x^2-1|+|2-x^2|=1 is -sqrt(2)<= x<=-1\lor 1<= x<= sqrt(2)

Solve for x: |x^2-1|+|2-x^2|=1

The solution is -sqrt(2)<= x<=-1\lor 1<= x<= sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

The solution is -sqrt(2)<= x<=-1\lor 1<= x<= sqrt(2)

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

- 2 \leq x \leq - 1 or 1 \leq x \leq 2

Solution steps

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

Find positive and negative intervals

x < - 2, - 2 \leq x \leq - 1, - 1 < x < 1, 1 \leq x \leq 2, x > 2

Solve the inequality for each interval

No Solution or - 2 \leq x \leq - 1 or 1 \leq x \leq 2

Merge Overlapping Intervals

- 2 \leq x \leq - 1 or 1 \leq x \leq 2

∣ x - 5 ∣ - ∣ 2 + x ∣ = 4

5 x^{2} - 3 = 2

3 x^{2} - 4 x^{2} - 1 + x - 1 = 0

∣ x - 2 ∣ + ∣ x - 5 ∣ = 3

∣ x - 1 ∣ - ∣ x ∣ + ∣ x + 1 ∣ = 2

What is the answer to |x^2-1|+|2-x^2|=1 ?
The answer to |x^2-1|+|2-x^2|=1 is -sqrt(2)<= x<=-1\lor 1<= x<= sqrt(2)
Solve for x: |x^2-1|+|2-x^2|=1
The solution is -sqrt(2)<= x<=-1\lor 1<= x<= sqrt(2)