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

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

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

The solution is x=(sqrt(17)-3)/2 \lor x=1

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

x = \frac{17 - 3}{2} or x = 1

Solution steps

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

Find positive and negative intervals

x \leq \frac{- 5 - 1}{2}, \frac{- 5 - 1}{2} < x < \frac{5 - 1}{2}, x \geq \frac{5 - 1}{2}

Solve the inequality for each interval

No Solution or x = \frac{17 - 3}{2} or x = 1

x = \frac{17 - 3}{2} or x = 1

x ∣ x ∣ = 4

∣ x + 1 ∣ + ∣ x - 1 ∣ = ∣ 2 x ∣

∣ 2 x + 5 ∣ + ∣ 2 x - 6 ∣ = 8

(- x^{2} + x - 1)^{3} + x^{2} - x + 1 = 0

∣ x - 1 ∣ ∣ x + 5 ∣ = 7

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