How do you find the roots, real and imaginary, of #y=4x^2 -7x -(x-2)^2 # using the quadratic formula?

1 Answer
Apr 13, 2016

Answer:

#x=(3+ sqrt(57))/6#
#x=(3- sqrt(57))/6#

Explanation:

Given -

#y=4x^2-7x-(x-2)^2#

Simplify

#y=4x^2-7x-(x^2-4x+4)#
#y=4^2-7x-x^2+4x-4#
#y=3x^2-3x-4#

It is in a quadratic form -

Then If #sqrt(b^2-(4ac))# is positive, its roots are real otherwise imaginary.

#sqrt((-3)^2-(4 xx 3 xx (-4))#
#sqrt(9-(-48)#
#sqrt(9+48)#
#sqrt(57)#

Since #sqrt(57)# is positive, the roots are real

#x=(-(-3)+-sqrt(57))/(2 xx 3)#

#x=(3+ sqrt(57))/6#
#x=(3- sqrt(57))/6#