What are the solutions to the equation s(s+1)(s+2)=1 ?

The solutions to the equation s(s+1)(s+2)=1 are s\approx 0.32471…

Find the zeros of s(s+1)(s+2)=1

The zeros of s(s+1)(s+2)=1 are s\approx 0.32471…

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s (s + 1) (s + 2) = 1

s \approx 0.32471 \dots

Solution steps

s (s + 1) (s + 2) = 1

Expand

s (s + 1) (s + 2) : s^{3} + 3 s^{2} + 2 s

s^{3} + 3 s^{2} + 2 s = 1

Move

1

to the left side

s^{3} + 3 s^{2} + 2 s - 1 = 0

Find one solution for

s^{3} + 3 s^{2} + 2 s - 1 = 0

using Newton-Raphson:

s \approx 0.32471 \dots

Apply long division:

\frac{s ^{3} + 3 s ^{2} + 2 s - 1}{s - 0.32471 \dots} = s^{2} + 3.32471 \dots s + 3.07959 \dots

s^{2} + 3.32471 \dots s + 3.07959 \dots \approx 0

Find one solution for

s^{2} + 3.32471 \dots s + 3.07959 \dots = 0

using Newton-Raphson:

No Solution for

s \in R

The solution is

s \approx 0.32471 \dots

x^{4} + 3 x^{2} - 2 = 0

(4 - y^{2})^{2} = 0

(2 x + 5)^{6} = (1 - x)^{6}

(3 x - 7)^{3} + 1 = 5

25 x^{2} (x + 3) - 9 (x + 3) = 0

What are the solutions to the equation s(s+1)(s+2)=1 ?
The solutions to the equation s(s+1)(s+2)=1 are s\approx 0.32471…
Find the zeros of s(s+1)(s+2)=1
The zeros of s(s+1)(s+2)=1 are s\approx 0.32471…