How do you find the roots, real and imaginary, of #y= 5x^2 - 6x - (2x+1)^2 # using the quadratic formula?

1 Answer
Oct 20, 2017

Answer:

Two Real roots at #x=5+-sqrt(26)#

Explanation:

#y=5x^2-6x-(2x+1)^2#

#y=5x^2-6x-(4x^2+4x+1)#

#rarr y=color(red)1x^2color(blue)(-10)xcolor(green)(-1)#

Applying the quadratic formula for roots:
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(green)c))/(2color(red)a)#
we have
#color(white)("XXX")x=(-color(blue)((-10))+-sqrt(color(blue)((-10))^2-4xxcolor(red)1xxcolor(green)((-1))))/(2xxcolor(red)1)#

#color(white)("XXX")x=(10+-sqrt(100+4))/2#

#color(white)("XXX")x=5+-sqrt(104)/2#

#color(white)("XXX")x=5+-sqrt(26)#