How do you find the roots, real and imaginary, of #y=-2x^2 + 7x +12(x/2-1)^2 # using the quadratic formula?

1 Answer
Nov 26, 2017

Answer:

#x_1=5/2+sqrt23/2i~~2.5+2.4i#
#x_2=5/2-sqrt23/2i~~2.5-2.4i#

When #5/2# in real and #sqrt23/2# is imaginary for both

Explanation:

for the form of:
#y=ax^2+bx+c#

the quadratic formula is:
#x_(1,2)=(-b+-sqrt{b^2-4*a*c})/(2*a)#


#y=-2x^2+7x+12(x/2-1)^2#

#=> y=-2x^2+7x+12(x^2/4+2*(x/2)*(-1)+1)#

#=> y=-2x^2+7x+12(x^2/4-x+1)#

#=> y=-2x^2+7x+3x^2-12x+12#

#=> y=x^2-5x+12#

#=>#
#a=1#
#b=-5#
#c=12#

#=>#
#x_(1,2)=(-b+-sqrt{b^2-4*a*c})/(2*a)=(-(-5)+-sqrt{25-4*1*12})/(2*1)=#
#=(5+-sqrt{25-48})/(2)=#
#=(5+-sqrt{-23})/(2)=#
#=(5+-sqrt23i)/(2)=#
#=>#
#x_1=(5+sqrt23i)/(2)=5/2+sqrt23/2i=2.5+sqrt23/2i~~2.5+2.4i#
#x_2=(5-sqrt23i)/(2)=5/2-sqrt23/2i=2.5-sqrt23/2i~~2.5-2.4i#