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

Aug 14, 2017

$4 \pm \frac{{\left(15\right)}^{\frac{1}{2}}}{2}$

#### Explanation:

If you expand the right side you should get
${x}^{2} - 8 x + 1$
Then just plug in to
$x = \frac{- b \pm {\left[{b}^{2} - 4 a c\right]}^{\frac{1}{2}}}{2 a}$
where a = 1 b = -8 c = 1

Aug 14, 2017

$x = 4 + \sqrt{15} ,$$4 - \sqrt{15}$

Refer to the explanation for the process.

#### Explanation:

Given:

$y = - 3 {x}^{2} - 4 x + {\left(2 x - 1\right)}^{2}$

FOIL (2x-1)^

$y = - 3 {x}^{2} - 4 x + \left[4 {x}^{2} - 4 x + 1\right]$

Gather like terms.

$y = \left(- 3 {x}^{2} + 4 {x}^{2}\right) - \left(4 x - 4 x\right) + 1$

Combine like terms.

$y = {x}^{2} - 8 x + 1$ $\leftarrow$ standard form: $y = a {x}^{2} + b x + c$,

where:

$a = 1$, $b = - 8$, and $c = 1$

Substitute $0$ for $y$ and solve for $x$ using the quadratic equation.

$0 = {x}^{2} - 8 x + 1$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in the given values.

$x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Simplify.

$x = \frac{8 \pm \sqrt{64 - 4}}{2}$

$x = \frac{8 \pm \sqrt{60}}{2}$

Prime factorize $60$.

$x = \frac{8 \pm \sqrt{2 \times 2 \times 3 \times 15}}{2}$

Simplify.

$x = \frac{8 \pm 2 \sqrt{15}}{2}$

Simplify.

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{8}^{4}}}} \pm \textcolor{red}{\cancel{\textcolor{b l a c k}{{2}^{1}}}} \sqrt{15}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{2}^{1}}}}}$

$x = 4 \pm \sqrt{15}$

Roots

$x = 4 + \sqrt{15} ,$$4 - \sqrt{15}$