What is the discriminent and the solutions of #2x^2+3x+5#?

1 Answer
Mar 7, 2018

Answer:

#x=-3/4+-sqrt(31)/4 i#

Explanation:

#color(blue)("Determining the discriminant") #

Consider the structure #y=ax^2+bx+c#

where #x=(-b+-sqrt(b^2-4ac))/(2a)#

The discriminant is the part #b^2-4ac#

So in this case we have:

#a=2; b=3 and c=5#

Thus the discriminant part #b^2-4ac -> (3)^2-4(2)(5) =-31#

As this is negative it means that the solution to #ax^2+bx+c# is such that #x# is not in the set of Real Numbers but is in the set of Complex numbers.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the solution for "ax^2+bx+c=0)#

#Using the above formula we have:

#x=(-3+-sqrt(-31))/4#

#x=-3/4+-sqrt(31xx(-1))/4#

#x=-3/4+-sqrt(31)/4 i#