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

1 Answer
Geoff K.
May 30, 2018

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).

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