1. To determine the roots of the equation s^3 + 7s^2 + 19s + 13 = 0, we can use various numerical methods or factorization. In this case, we can observe that s = -1 is a root of the equation by substituting it:
(-1)^3 + 7(-1)^2 + 19(-1) + 13 = -1 + 7 - 19 + 13 = 0
By performing polynomial division, we find that the equation can be factored as:
(s + 1)(s^2 + 6s + 13) = 0
Using the quadratic formula to solve the quadratic equation, we find the roots:
s = (-6 ± √(6^2 - 4(1)(13))) / (2(1))
s = (-6 ± √(-8)) / 2
s = (-6 ± 2i√2) / 2
s = -3 ± i√2
Therefore, the roots of the equation are s = -1, -3 + i√2, and -3 - i√2.
2. To determine the roots of the equation s^3 + s = 1, we can rearrange it as s^3 + s - 1 = 0. Since this is a cubic equation, finding exact solutions can be challenging. We can use numerical methods or approximation techniques to find the roots. One approach is to use the Newton-Raphson method or a calculator or software that can solve cubic equations. The approximate roots for this equation are:
s ≈ 0.68233
s ≈ -1.68233
s ≈ 0.00000
3. To determine the roots of the equation s^5 + 105s^4 + 510s^3 + 1010s^2 + 1004s + 400 = 0, again, finding exact solutions can be difficult for higher degree equations. We can use numerical methods or approximation techniques. Using a calculator or software, we find the approximate roots for this equation:
s ≈ -6.72976
s ≈ -4.14184
s ≈ -1.40492
s ≈ -0.196644 + 0.69586i
s ≈ -0.196644 - 0.69586i
These are the approximate roots of the equation.