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

1 Answer
Aug 23, 2017

Answer:

#x=(3+sqrt57)/6,# #(3-sqrt57)/6#

Explanation:

Solve:

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

Expand #(x-2)^2#.

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

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

Simplify.

#y=4x^2-7x-x^2+4x-4#

Gather like terms.

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

Combine like terms.

#y=3x^2-3x-4#

To solve for roots, substitute #0# for #y#. Then solve for #x# using the quadratic formula.

#0=3x^2-3x-4#

Quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the given values.

#x=(-(-3)+-sqrt((-3)^2-4*3*-4))/(2*3)#

Simplify.

#x=(3+-sqrt(9+48))/6#

#x=(3+-sqrt57)/12# #larr# #57# is the product of two prime numbers.

#x=(3+sqrt57)/6,# #(3-sqrt57)/6#