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

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