Question 2823
>>When are there two solutions in quadratic equations???<<
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<PRE>
Every quadratic equation

{{{ax^2 + bx + c = 0}}}

has either

1. Two real solutions
2. One real solution
3. Two conjugate imaginary solutions

The quadratic equation:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}  

has two real solutions when the " ± " part gives two different 
answers, one for the + and another for the -.  This will always 
be the case when the radicand, or "discriminant",

{{{b^2 - 4ac}}}<font face = "symbol">¹</font>0 

If 

{{{b^2 - 4ac}}} > 0, the two solutions are real, and when

{{{b^2 - 4ac}}} < 0, the two solutions are conjugate imaginary
numbers.

However, if 

{{{b^2 - 4ac}}} = 0, there is but one real solution, because the
"±" part is "±0", and whether we add 0 or subtract 0, we still 
get the same number.
</PRE>
Edwin <font face = "wingdings" color = "orange" size = 7>J</font>