How do you find the roots, real and imaginary, of #y=(-2x+4)(x+1)+3x^2+2x # using the quadratic formula?

1 Answer
May 30, 2018

Answer:

The root is #x=–2,# with multiplicity two.

Explanation:

Convert the right-hand side to the form #ax^2+bx+c#.

#y="   "(–2x+4)(x+1)"  " + 3x^2 + 2x#
#color(white)y=–2x^2-2x+4x+4 + 3x^2 + 2x#
#color(white)y=x^2+4x+4#

The roots of this function occur when #y=0#. We replace #y# with #0# and solve for #x:#

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

Using the quadratic formula, the solutions for a quadratic #ax^2 + bx + c=0# are

#x=(–b+-sqrt(b^2-4ac))/(2a)#

For the given quadratic, #a = 1, b=4, and c = 4#. We plug these values into the quadratic formula as follows:

#x=(–4+-sqrt(4^2-4(1)(4)))/(2(1))#

#x=(–4+-sqrt(16-16))/2#

#x=(–4+-sqrt(0))/2#

#x=(–4)/2" " = –2#

But wait—shouldn't we get two roots?
Yes, we should, and we did! Since the discriminant #b^2-4ac# was equal to 0, the "one" root we got actually has a multiplicity of two. Meaning, the factor #x+2# appears twice in the factorization of #x^2+4x+4.#

Indeed, we can confirm this if we factor #x^2+4x+4# the shorter way, where we find two numbers that add to 4 and multiply to 4; those two numbers are 2 and 2, giving us

#x^2+4x+4 = (x+2)(x+2).#