What is the answer to |x^2-3x-1|=1 ?

The answer to |x^2-3x-1|=1 is x=(3-sqrt(17))/2 \lor x=0\lor x=3\lor x=(3+sqrt(17))/2

Solve for x: |x^2-3x-1|=1

The solution is x=(3-sqrt(17))/2 \lor x=0\lor x=3\lor x=(3+sqrt(17))/2

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x^{2} - 3 x - 1 = 1

x = \frac{3 - 17}{2} or x = 0 or x = 3 or x = \frac{3 + 17}{2}

Decimal Notation

x = - 0.56155 \dots, x = 0, x = 3, x = 3.56155 \dots

Solution steps

x^{2} - 3 x - 1 = 1

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

x^{2} - 3 x - 1 = - 1 or x^{2} - 3 x - 1 = 1

x^{2} - 3 x - 1 = - 1 or x^{2} - 3 x - 1 = 1

(x = 0 or x = 3) or (x = \frac{3 - 17}{2} or x = \frac{3 + 17}{2})

x = \frac{3 - 17}{2} or x = 0 or x = 3 or x = \frac{3 + 17}{2}

5 ∣ 5 x - 7 ∣ - 1 = 14

5 - x^{2} = 3 - x

- 6 = ∣ 3 x - 3 ∣

- 3 ∣ x - 5 ∣ + 4 = 13

∣ y ∣ = 16, x + y = 0

What is the answer to |x^2-3x-1|=1 ?
The answer to |x^2-3x-1|=1 is x=(3-sqrt(17))/2 \lor x=0\lor x=3\lor x=(3+sqrt(17))/2
Solve for x: |x^2-3x-1|=1
The solution is x=(3-sqrt(17))/2 \lor x=0\lor x=3\lor x=(3+sqrt(17))/2