What is the answer to |x-1|+1=x^2-4x ?

The answer to |x-1|+1=x^2-4x is x=(3-sqrt(17))/2 \lor x=5

Solve for x: |x-1|+1=x^2-4x

The solution is x=(3-sqrt(17))/2 \lor x=5

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∣ x - 1 ∣ + 1 = x^{2} - 4 x

x = \frac{3 - 17}{2} or x = 5

Decimal

x = - 0.56155 \dots, x = 5

Solution steps

∣ x - 1 ∣ + 1 = x^{2} - 4 x

Find positive and negative intervals

x < 1, x \geq 1

Solve the inequality for each interval

x = \frac{3 - 17}{2} or x = 5

∣ 6 x - 3 ∣ = 33

∣ x - 3 ∣ + ∣ x - 6 ∣ = 5

∣ 11 - 2 x ∣ + 6 = 10

8 - ∣ 1 - 2 x ∣ = x

4 ∣ 3 x + 1 ∣ + 6 = 18

What is the answer to |x-1|+1=x^2-4x ?
The answer to |x-1|+1=x^2-4x is x=(3-sqrt(17))/2 \lor x=5
Solve for x: |x-1|+1=x^2-4x
The solution is x=(3-sqrt(17))/2 \lor x=5