What is the answer to |9-x^2|=x-1 ?

The answer to |9-x^2|=x-1 is x=(sqrt(41)-1)/2 \lor x=(1+sqrt(33))/2

Solve for x: |9-x^2|=x-1

The solution is x=(sqrt(41)-1)/2 \lor x=(1+sqrt(33))/2

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9 - x^{2} = x - 1

x = \frac{41 - 1}{2} or x = \frac{1 + 33}{2}

Decimal Notation

x = 2.70156 \dots, x = 3.37228 \dots

Solution steps

9 - x^{2} = x - 1

Find positive and negative intervals

x < - 3, - 3 \leq x \leq 3, x > 3

Solve the inequality for each interval

No Solution or x = \frac{41 - 1}{2} or x = \frac{1 + 33}{2}

x = \frac{41 - 1}{2} or x = \frac{1 + 33}{2}

∣ 3 x - 15 ∣ - 5 = - 2

- 4 ∣ 2 x - 8 ∣ = 9

2 x^{3} + 8 - 7 = 11

\frac{1}{2} x - 3 = 0

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

What is the answer to |9-x^2|=x-1 ?
The answer to |9-x^2|=x-1 is x=(sqrt(41)-1)/2 \lor x=(1+sqrt(33))/2
Solve for x: |9-x^2|=x-1
The solution is x=(sqrt(41)-1)/2 \lor x=(1+sqrt(33))/2