Explanation:

Let’s go through each statement and see what’s really happening with this quadratic:

- Statement 1: *One of the roots is always less than 1 if a, b, c are all positive.*

- Plug in x = 1 into the quadratic:

(a+b+c)(1)^2 - (2a+2b)(1) + (a+b) = a+b+c - 2a - 2b + a + b = c

So, f(1) = c (which is positive).

- For x very large negative, the quadratic (since leading coefficient is positive) goes up to infinity, so on the far left, f(x) is positive.

- So, both at x = 1 (f(1) = c > 0) and x = -8 (f(x) ? +8), value is positive, which means there isn't necessarily a root less than 1.

- Counterexample: a = 1, b = 2, c = 3 gives quadratic: 6x² - 6x + 3 = 0. Roots are both greater than 1.

- So, Statement 1 is actually not always correct.

- Statement 2: *One of the roots is always negative if a, b, c are all negative.*

- Let’s make a, b, c negative values.

- Now leading coefficient (a + b + c) is negative, so the parabola opens downward.

- Plugging x = 0: f(0) = a + b (which is negative), at x = 0, value is negative.

- For x = -8, leading coefficient negative, so f(x) ? -8 as x ? -8.

- No guarantee for a negative root—try a = -1, b = -2, c = -3:

Equation: (-6)x² - (-6)x + (-3) = -6x² + 6x -3 = 0. Both roots are positive.

- So, Statement 2 is not always correct.

- Option 1: 1 only

- Option 2: 2 only

- Option 3: Both 1 and 2

- Option 4: Neither 1 nor 2

Here’s the deal: Neither statement is always true. Both can be shown false with real, simple examples.