Question

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

Answer 1

Answer:

a < - $\frac{2}{3}$

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 < $\frac{8}{-12}$, that is

a < - $\frac{2}{3}$