What is a so the polynomial p(x)= (a+1)x^2 + 2ax + (a+2) has two roots?

a < -

Step-by-step explanation:

Using the discriminant Δ = b² - 4ac

For a polynomial to have 2 real roots

b² - 4ac > 0

Given

p(x) = (a + 1)x² + 2ax + (a + 2)

with a = a + 1, b = 2a and c = a + 2, then

(2a)² - 4(a + 1)(a + 2) > 0 ← expand and simplify left side

4a² - 4(a² + 3a + 2) > 0

4a² - 4a² - 12a - 8 > 0

- 12a - 8 > 0 ( add 8 to both sides )

- 12a > 8

Divide both sides by - 12, reversing the symbol as a result of dividing by a negative quantity

a < , that is

a < -

25/4

step-by-step explanation:

call segment cd's length x. because dc and ab are parallel, angle acd and cab, so triangles adc and abc are similar and we can set up the equation x / ac = ac / ab, or x / 5 = 5 / 4 and x = 25/4.

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!