Since each of the questions being asked in a), b), and c) keep #c = 0# out of the realm of possibility, we can safely assume for this problem that #c != 0#. Therefore, we can begin by removing the fraction:
#(3x)/(x^2+1) = c#
#3x = c(x^2+1)#
#3x = cx^2 + c#
#cx^2 - 3x + c = 0#
This is a quadratic equation, now in #ax^2+bx+c# format (this #c# is not the same as the #c# in our problem...poor luck on the naming, there...), which means we can examine the Quadratic Formula to determine what to expect for various values of #c#:
#x = (-b +- sqrt(b^2-4ac))/(2a) = (-(-3) +- sqrt((-3)^2-4(c)(c)))/(2(c))#
# = (3 +- sqrt(9-4c^2))/(2c)#
Let's treat each question in turn:
#bb(c > 3/2):#
In this case, since #c > 3/2# it follows that #c^2 > 9/4#, and further follows that #4c^2 > 9#. This means the discriminant in the formula above (the #b^2-4ac# portion) would be negative. A negative discriminant indicates that the equation would have no real roots; the only roots for the problem would be a pair of complex roots (one complex root and its conjugate, actually).
#bb(c = 3/2):#
In this case, since #c = 3/2#, we can follow this logic:
#c = 3/2 => c^2 = 9/4 => 4c^2 = 9 => sqrt(9-4c^2) = sqrt(0) = 0#
Since the radical portion evaluates to 0, we will end up with a single real root (technically of multiplicity 2, meaning it is a double root.)
#bb(0 < c < 3/2):#
In this case, since #0 < c < 3/2#, we know that #c^2 < 9/4#, or that #4c^2 < 9#. This means the radical term #9-4c^2# will be a positive value, and thus we will end up with two different real roots to the overall equation.
