I'm having trouble understanding how one would construct a direct proof for the following statement:
"The quadratic equation $ax^2+bx+c = 0$ has a solution whenever $b^2\ge 4ac$."
I understand that the statement is true overall, and that it is also equivalent to its converse. Would one have to take a divide-by-case approach where we first prove the statement when $b^2 = 4ac$, and then show when $b^2 \gt 4ac$?
I've attached a screenshot of a proof which attempts to prove the equivalence between the original statement and its converse:
For my second question, the proof works its way backwards from the quadratic equation itself to meet the initial condition, but how would one do the reverse?
Additionally, I'm confused as to how the writer came up with the idea to factor the equation and introduce additional element to both sides to reconstruct the quadratic formula. How does one come up with such an idea to construct a proof?