Question 74863
A)
When {{{b^2-4ac<0}}} there are no real solutions because if you have {{{sqrt(x)}}} and x is a negative number, you wont have a real answer. You will have a complex solution. The reason why is because {{{x^2}}} is always positive.



B)
When b^2-4ac is equal to zero, there is one solution because if you have
{{{(6+-sqrt(0))/2}}} you get {{{(6+-0)/2}}} and since zero has no sign, there is only one answer which in this case it's 3



D)
When b^2-4ac is equal to a perfect square (ie 9), there are two rational solutions because a perfect square is a rational number. For instance, if we have
{{{(6+-sqrt(9))/2}}} it equals {{{(6+3)/2}}} and {{{(6-3)/2}}} which are both rational numbers. Note: this is the same as part C, since any number (such as 2) is rational (ie {{{2=2/1}}}) so any solution (that isn't irrational or complex) is rational



E)
When b^2-4ac is not equal to a perfect square (like 27), there are two irrational solutions because the square root of any number that is not a perfect square is irrational. Since an irrational number added to a rational number gives you an irrational number, the solutions will be irrational. For example, lets say we have:
{{{(6+-sqrt(27))/2}}}} which becomes {{{(6+sqrt(27))/2=8.59807621135331
}}} and {{{(6-sqrt(27))/2=3.40192378864669
}}} which are both irrational (cannot be represented as a fraction)