Any other solution?

Okay so this problem is a bit tricky, I have few comments and the feedback is most appreciated.

My first notice is the importance of the conditions given, so we know the (a) > 0 , (b) < 0.

Moreover, we see in the equation given that: x^2 - ax +b = 0

Considering the first condition we see that the equation is really in this shape: x^2 - ax - b ; because b is a negative number.

If their sum in the equation is negative then after we factor them and solve they are going to change signs and thus their sum after becoming the solutions is a positive sum.

And the tricky part regarding their multiplication is irrelevant, they would never equal to -b because their multiplication is going to be definitely negative( they have opposite signs) and b < 0; so -b == +ve value.

It becomes much clearer as you demonstrate it on a simple equation as mentioned in the above comments.

I found it helpful to think about a real-life example. I know that since b is negative, the two roots need to have opposite signs (per factoring). And since a is positive, in the equation it will flip to negative with the minus sign, so when I factor, I want the number I subtract to be greater than the number I add.

For example, (x-5)(x+3) = x^2 -2x -15 would meet this criteria, since this creates a negative for both -a and b (a=2, b=-15).

If I look at this test case, I see that the roots that solve the equation are x=5 and x=-3.

Then I just checked these against the criteria and logically deduced why that would be generalizable.

A: They have opposite signs. YES --> We already knew this would have to be the case in order to factor into a number less than zero for b.

B: Their sum is greater than zero: YES --> We see this in our example. Thinking more generally, we know this has to be true in order to create the situation where we could subtract a number for the second term (the one that goes with "x"), as in our equation.

C: Their products equal -b: NO --> In our case, their products equal b, not negative b. We also know from the general rules of factoring that the products of a negative and positive root will be the same as the negative value at the end of our equation.

So the answer is A and B.

still unsure on option a

The given quadratic equation is $\(x^2-a x+b=0\)$.
This equation is in the standard form $\(A x^2+B x+C=0\)$, where:
- $A=1$
- $B=-a$
- $C=b$

Let the two roots of the equation be $\(x_1\)$ and $\(x_2\)$.


1. Sum of the roots $\(\left(x_1+x_2\right):-B / A=-(-a) / 1=a\)$
2. Product of the roots $\(\left(x_1 \cdot x_2\right): C / A=b / 1=b\)$

We are given the conditions:
- $\(a>0\)$
- $\(b<0\)$

Now let's evaluate each statement:

Statement 1: They have opposite signs.
- This refers to the product of the roots $\(\left(x_1 \cdot x_2\right)\)$.
- From Vieta's formulas, $\(x_1 \cdot x_2=b\)$.
- We are given that $\(b<0\)$.
- For the product of two numbers to be negative, one number must be positive and the other must be negative.
- Therefore, the roots have opposite signs.
- This statement is TRUE.

Statement 2: Their sum is greater than zero.
- This refers to the sum of the roots $\(\left(x_1+x_2\right)\)$.
- From Vieta's formulas, $\(x_1+x_2=a\)$.
- We are given that $\(a>0\)$.
- Therefore, the sum of the roots is greater than zero.
- This statement is TRUE.

Statement 3: Their product equals \(-b\).
- This refers to the product of the roots $\(\left(x_1 \cdot x_2\right)\)$.
- From Vieta's formulas, $\(x_1 \cdot x_2=b\)$.
- The statement claims the product equals $\(-b\)$.
- Since $\(b<0\)$, then $-b$ would be a positive value. For example, if $\(b=-5\)$, then $\(-b=5\)$.
- The product is $b$, not $-b$ (unless $b=0$, which is excluded by $b<0$ ).
- Therefore, this statement is FALSE.