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

The answer to |x^2+x-1|=1 is x=-2\lor x=-1\lor x=0\lor x=1

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

The solution is x=-2\lor x=-1\lor x=0\lor x=1

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x^{2} + x - 1 = 1

x = - 2 or x = - 1 or x = 0 or x = 1

Solution steps

x^{2} + x - 1 = 1

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

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

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

(x = - 1 or x = 0) or (x = - 2 or x = 1)

x = - 2 or x = - 1 or x = 0 or x = 1

\frac{2}{3} x + \frac{3}{2} (x - 5) \leq x

x^{4} - 2 x^{3} + 27 x - 54

x (2 x + 4) = 0

2 x + 12 \leq 2 x + 12

8 - 2 x \leq 2

What is the answer to |x^2+x-1|=1 ?
The answer to |x^2+x-1|=1 is x=-2\lor x=-1\lor x=0\lor x=1
Solve for x: |x^2+x-1|=1
The solution is x=-2\lor x=-1\lor x=0\lor x=1