Consider the polynomial p(s) = s2 + bs + c where b and c are real numbers. Show that all the roots of p(s) are both contained in the open left half plane {s : s < 0} if and only if b > 0 and c > 0. Hint: use the quadratic formula.

It is shown in the explanation

Step-by-step explanation:

p(s) = s² + bs + c

a = 1

b = b

c = c

We get Δ as follows

Δ = (b²-4*a*c) = b² - 4*1*c = b² - 4c > 0  ⇒   b² > 4c  ⇔ c > 0

s = (-b + √(b² - 4c))/2(1)

⇒   s₁ = (-b + √(b² - 4c))/2

s₂ = (-b - √(b² - 4c))/2(1)

⇒   s₂ = (-b - √(b² - 4c))/2

We have that -b < 0  ⇔ b > 0

then s₁ < 0 and s₂ < 0 ⇔ c > 0 and b > 0

For roots to lie on the left half plane, b ⊃ 0 and c ⊃0

Step-by-step explanation:

From quadratic formula, we have;

x = -b±√(b²-4ac)/2a

From the given expression p(s) = s² + bs + c,

x = s

a = 1

b = b

c = c

The quadratic formula can then be written as;

s =  -b±√(b²-4*1*c)/2*1

   =  -b±√(b²-4c)/2

s₁ =  -b+√(b²-4c)/2

s₂ =  -b±√(b²-4c)/2

From the equation above,

Sum of root = -b

Product of root = c

If both the root lie on left side of the s-plane, then sum of roots will be negative. Hence, -b ∠0. That is, b ⊃0

Also, the product root will be positive, c ⊃ 0

Hence, for roots to lie on the left half plane, b ⊃ 0 and c ⊃0

d

gimmee one of those gold crowns!

~coco

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