My strategy has been to write an example of second degree equation that is easy to solve as x^2-x-4 = 0, where a = 1 and b = -4. This is easily factorized as (x-2)(x+1) = 0 so that the solutions are 2 and -1. Thus, we can screening our choices:

❑ They have opposite signs: 2 and -1 have opposite signs -> CORRECT

❑ Their sum is greater than zero: 2 + (-1) = 1 >0 -> CORRECT

❑ Their product equals –b: 2*(-1) = -2 while b = -4 -> WRONG

Answers are A and B!

for equation ax^2+bx+c = 0
sum of roots = -b/a
product of roots = c/a

In case of this question:
x^2-ax+b = 0
sum of roots = a (since a> 0 , option b is correct)
product of roots = b (option c incorrect)

product of roots = b and since b is less than 0 , therefore both roots should be of opposite sign (option a correct)

Any other solution?
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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.

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