What is the answer to 1|x^2-x|+|x+1|=5 ?

The answer to 1|x^2-x|+|x+1|=5 is x=1-sqrt(7)\lor x=2

Solve for x: 1|x^2-x|+|x+1|=5

The solution is x=1-sqrt(7)\lor x=2

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1 x^{2} - x + ∣ x + 1 ∣ = 5

x = 1 - 7 or x = 2

Solution steps

1 \cdot x^{2} - x + ∣ x + 1 ∣ = 5

Find positive and negative intervals

x < - 1, - 1 \leq x \leq 0, 0 < x < 1, x \geq 1

Solve the inequality for each interval

x = 1 - 7 or No Solution or x = 2

x = 1 - 7 or x = 2

∣ 3 x + 2 ∣ = x + 1, ∣ 2 x - 3 ∣ = x - 5, ∣ 3 x - 2 ∣ = 2 x + 3

y, 17 x^{2} - 16 ∣ x ∣ y + 17 y^{2} = 225

x^{2} + 2 ∣ x ∣ = 15

x^{2} + x + 4 = 0

x^{2} - 4 x + x^{2} - 5 - 1 = 0

What is the answer to 1|x^2-x|+|x+1|=5 ?
The answer to 1|x^2-x|+|x+1|=5 is x=1-sqrt(7)\lor x=2
Solve for x: 1|x^2-x|+|x+1|=5
The solution is x=1-sqrt(7)\lor x=2