
Michael F. answered 02/15/15
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Hello Meredith,
The following is a break down to the solution of your problem....
Here are the scenarios for all discriminants and their results.
1. If b² – 4ac = 0 then there are two equal roots and rational
2. If b² – 4ac > 0 and a perfect square, then the solutions are rational and different
3. If b² – 4ac > 0, but not a perfect square, then the solutions are irrational and different
4. If b² – 4ac < 0 then both solutions are complex conjugates
2. If b² – 4ac > 0 and a perfect square, then the solutions are rational and different
3. If b² – 4ac > 0, but not a perfect square, then the solutions are irrational and different
4. If b² – 4ac < 0 then both solutions are complex conjugates
So....
since the vertex is in the 3rd quadrant with a y intercept of -6, I know that the roots will cross the x-axis
The first choice is -11. This can not happen because this reverts back to situation number 4 and imaginary roots do not cross the x-axis and I know this parabola has real roots.
The second choice of 0 does not work because this reverts to situation number 1 which yields two equal roots and rational. This can not happen because since the vertex is at (-1.2, -9.1) you can tell that there are two different roots and they are not equal.
The third choice of 25 does not work because this reverts to situation number 2 which yields rational roots.
This leaves you with choice 4 because a discriminant of a positive non perfect square will give two irrational roots.
I hoped this helped.