What is the answer to |0.127x+9|=x^2-2 ?

The answer to |0.127x+9|=x^2-2 is x=(127-9sqrt(543409))/(2000)\lor x=(127+9sqrt(543409))/(2000)

Solve for x: |0.127x+9|=x^2-2

The solution is x=(127-9sqrt(543409))/(2000)\lor x=(127+9sqrt(543409))/(2000)

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∣ 0.127 x + 9 ∣ = x^{2} - 2

x = \frac{127 - 9 543409}{2000} or x = \frac{127 + 9 543409}{2000}

Decimal Notation

x = - 3.25373 \dots, x = 3.38073 \dots

Solution steps

∣ 0.127 x + 9 ∣ = x^{2} - 2

Find positive and negative intervals

x < - \frac{9000}{127}, x \geq - \frac{9000}{127}

Solve the inequality for each interval

No Solution or (x = \frac{127 - 9 543409}{2000} or x = \frac{127 + 9 543409}{2000})

x = \frac{127 - 9 543409}{2000} or x = \frac{127 + 9 543409}{2000}

∣ x + 7.5 ∣ = - 2

∣ x + 1 ∣ = x^{2} + x

∣ x ∣ = x - 6

∣ - x + 5 ∣ + 4 = 9

∣ x + 1 ∣ = x^{2} + 5

What is the answer to |0.127x+9|=x^2-2 ?
The answer to |0.127x+9|=x^2-2 is x=(127-9sqrt(543409))/(2000)\lor x=(127+9sqrt(543409))/(2000)
Solve for x: |0.127x+9|=x^2-2
The solution is x=(127-9sqrt(543409))/(2000)\lor x=(127+9sqrt(543409))/(2000)