How do you find the roots, real and imaginary, of #y=2x^2 + 13x + 6+4(x -1)^2 # using the quadratic formula?

1 Answer
Dec 29, 2015

Answer:

#x = (-5 + sqrt(-215))/12# or #x = (-5 - sqrt(-215))/12#

Explanation:

First expand and simplify to ensure just one term for each power of x.
#y = 2x^2 +13x +6 +4(x^2 -2x + 1)#
#y = 2x^2 +13x + 6 + 4x^2 -8x +4#
#y=6x^2 +5x +10#
Then use #x = (-b +- sqrt(b^2 - 4ac))/(2a)#
#x = (-5 +- sqrt( 5^2 - 4*6*10))/(2*6)#
#x = (-5 +-sqrt(25 - 240))/12#
#x = (-5 +- sqrt(-215))/12#
There are only imaginary roots to this equation, meaning that in graphical terms the graph never intersects the x axis.