Given Information:

CD are the zeros of the quadratic equation x² - 10x - 11b.

alpha and beta are the zeros of the quadratic equation x² - 10x - 11b.


To Find:

The value of alpha x beta x C x b.

Solution:

Let's first find the sum and product of the zeros of the quadratic equation x² - 10x - 11b.


The sum and product of the roots of the quadratic equation ax² + bx + c = 0 are given by the following formulas:

Sum of the roots = - b/a

Product of the roots = c/a


The quadratic equation x² - 10x - 11b can be written as x² - (CD + alpha + beta)x + alpha beta = 0.

Comparing with ax² + bx + c = 0, we get

a = 1

b = -(CD + alpha + beta) = -10

c = alpha beta = -11b


Sum of the roots = -b/a = (CD + alpha + beta)/1 = CD + alpha + beta

Product of the roots = c/a = (alpha beta)/1 = -11b


Therefore, we have the following two equations:

CD + alpha + beta = 10 ...(1)

alpha beta = -11b ...(2)



We know that the product of the zeros of a quadratic equation ax² + bx + c = 0 is given by c/a.

Therefore, the product of the zeros of x² - 10x - 11b is -11b/1 = -11b.


We know that if alpha and beta are the roots of a quadratic equation ax² + bx + c = 0, then the equation can be written as a(x - alpha)(x - beta) = 0.

Expanding this equation, we get ax² - a(alpha + beta)x + a(alpha beta) = 0.

Comparing with x² - 10x - 11b, we get

a = 1

alpha + beta = 10 + CD

alpha beta = -11b


Therefore, the given quadratic equation can be written as (x - alpha)(x - beta) = x² - (alpha + beta)x + alpha beta = 0.

Comparing with ax² + bx + c = 0, we get

a = 1

b = -(alpha + beta) = -(CD + 10)

c = alpha beta = -11b


Therefore, C is equal to