For this question, could it be possible that the equation in Quantity A has no solution? Or only two? For example, wouldn’t it be possible that b^2+4a(c-7) < 0?
I don’t think so - the absolute value makes this difficult.
Can you please explain why it’s not possible for there to be zero solutions?
We have b^2 - 4ac - 28a and b^2 - 4ac + 28a, for there to exist solutions to both equations, b^2 - 4ac must not be less than -28a or not less than 28a - so to fulfil solutions to both equations, b^2-4ac should be >= 28a. How do we know this guaranteed if the only info given is a,b,c != 0? How does the absolute value guarantee an existence of solution to the quadratic?
I’m not sure where 28a comes from? Also to answer your question, think of an example graph.
(Edit: I indeed made a mistake, see the response below. You can have 0 solutions; eg |x^2 + 4x + 14|)
The quantity A is a quadratic with absolute value applied on it. A quadratic equation ax² + bx + c = 0 can have 0, 1 or 2 roots depending on the discriminant value (remember that roots are points where the polynomial cuts the x axis). After applying the modulus it just flips the negative part to positive. However it is immaterial to the question at hand, how many roots does the quadratic have, as the solutions to |ax² + bc + c| = 7 are the points where this left hand side graph intersects y = 7. That could be 0, 2, 3 or 4 depending on where they intersect. Keep a clarity in mind between roots and solutions (imagine how different ways a horizontal line — can intersect a " W " figure)