How do you find the roots, real and imaginary, of #y=-(x+5)^2+2x^2 + 5x - 12 # using the quadratic formula?

1 Answer
Jul 11, 2017

Answer:

#x_1~~6.437+0i#
#x_2~~-1.437+0i#

Explanation:

Quadratic formula, given by:
#(-b+-sqrt(b^2-4ac))/(2a)#

Before we proceed, let's us expand the equation of #y#
#y=-(x+5)^2+2x^2+5x-12#
#y=-(x^2+10x+25)+2x^2+5x-12#
#y=-x^2-10x-25+2x^2+5x-12#
#y=(2-1)x^2+(5-10)x+(-37)#
#y=(1)x^2+(-5)x+(-37)#

Determine the #a#,#b#, and #c#
#a=1#
#b=-5#
#c=-37#

Use the formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(-5)+-sqrt((-5)^2-4(1)(-37)))/(2(1))#
#x=(5+-sqrt(25+37))/2#
#x=(5+-sqrt(62))/2#

#x_1=(5+sqrt(62))/2~~6.437#
#x_2=(5-sqrt(62))/2~~-1.437#

In this case, the roots are real, their imaginary parts are 0
#x_1~~6.437+0i#
#x_2~~-1.437+0i#