How do you find the roots, real and imaginary, of #y=(2x-4)(x-1)-5x^2+4x # using the quadratic formula?

1 Answer
Aug 15, 2017

Answer:

#x=-(1+sqrt13)/3,##-(1-sqrt13)/3#

Refer to the explanation for the process.

Explanation:

Given:

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

FOIL #(2x-4)(x-1)#.

#y=[2x^2-2x-4x+4]-5x^2+4x#

Simplify.

#y=[2x^2-6x+4]-5x^2+4x#

Gather like terms.

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

Combine like terms.

#y=-3x^2-2x+4# #larr# quadratic equation standard form:

#y=ax^2+bx+c#,

where:

#a=-3#, #b=-2#, and #c=4#.

Quadratic Formula

Substitute #0# for #y#. Solve for #x#.

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

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

Plug in the known values.

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

Simplify.

#x=(2+-sqrt(4+48))/(-6)#

#x=(2+-sqrt52)/(-6)#

Prime factorize #52#.

#x=(2+-sqrt(2xx2xx13))/(-6)#

Simplify.

#(2+-2sqrt13)/(-6)#

Simplify.

#(color(red)cancel(color(black)(2^1))+-(color(red)cancel(color(black)(2^1))sqrt13))/(-(color(red)cancel(color(black)(6^3)))#

#x=(1+-sqrt3)/-3#

Roots

#x=-(1+sqrt13)/3,##-(1-sqrt13)/3#