How do you find the zeros, real and imaginary, of #y=19x^2-4x-3 #using the quadratic formula?

1 Answer
Mar 27, 2018

Answer:

#x=(2+sqrt(41))/(19)# and #x=(2-sqrt(41))/(19)#.

Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a)#. #a#,#b#, and #c# come from #ax^2+bx+c#, the standard form of a quadratic equation.

The equation is already in standard form, so we can simply substitute and simplify!

#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(-4)+-sqrt((-4)^2-4(19)(-3)))/(2(19))#
#x=(4+-sqrt(16-(-228)))/(38)#
#x=(4+-sqrt(244))/(38)#
244 is not a perfect square, so we could stop here. However, notice that 244 is divisible by 4, which is a perfect square, so we can simplify #sqrt(244)#!
#x=(4+-sqrt(4*41))/(38)#
#x=(4+-2sqrt(41))/(38)#
We can pull a 2 out of each term and simplify:
#x=(cancel2(2+-sqrt(41)))/(cancel2(19))#
#x=(2+-sqrt(41))/(19)#
So our solutions are #x=(2+sqrt(41))/(19)# and #x=(2-sqrt(41))/(19)#.