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

May 30, 2018

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

#### Explanation:

Convert the right-hand side to the form $a {x}^{2} + b x + c$.

$y = \text{ "(–2x+4)(x+1)" } + 3 {x}^{2} + 2 x$
color(white)y=–2x^2-2x+4x+4 + 3x^2 + 2x
$\textcolor{w h i t e}{y} = {x}^{2} + 4 x + 4$

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

$0 = {x}^{2} + 4 x + 4$

Using the quadratic formula, the solutions for a quadratic $a {x}^{2} + b x + c = 0$ are

x=(–b+-sqrt(b^2-4ac))/(2a)

For the given quadratic, $a = 1 , b = 4 , \mathmr{and} c = 4$. 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 ${b}^{2} - 4 a c$ 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 ${x}^{2} + 4 x + 4.$

Indeed, we can confirm this if we factor ${x}^{2} + 4 x + 4$ 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

${x}^{2} + 4 x + 4 = \left(x + 2\right) \left(x + 2\right) .$