# How do you find the roots, real and imaginary, of y= 2x^2 + 7x+(2x- 1 )^2  using the quadratic formula?

Jul 2, 2017

$x = \frac{- 3 + i \sqrt{15}}{12} ,$$\frac{- 3 - i \sqrt{15}}{12}$

#### Explanation:

Solve:

$y = 2 {x}^{2} + 7 x + {\left(2 x - 1\right)}^{2}$

Expand ${\left(2 x - 1\right)}^{2}$ using the square of a difference: $\left({a}^{2} - {b}^{2}\right) = {a}^{2} - 2 a b + {b}^{2}$, where $a = 2 x$ and $b = 1$.

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

Simplify.

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

Add the result to the rest of the equation.

$y = 2 {x}^{2} + 7 x + 4 {x}^{2} - 4 x + 1$

Simplify.

$y = 2 {x}^{2} + 4 {x}^{2} + 7 x - 4 x + 1$

$y = 6 {x}^{2} + 3 x + 1$

Substitute $0$ for $y$.

$0 = 6 {x}^{2} + 3 x + 1$ is a quadratic equation in standard form: ${a}^{2} + b x + c$, where $a = 6$, $b = 3$, and $c = 1$.

Use the quadratic formula to solve for the roots (values for $x$).

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

Substitute the values for $a , b , \mathmr{and} c$ into the formula.

$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 6 \cdot 1}}{2 \cdot 6}$

Simplify.

$x = \frac{- 3 \pm \sqrt{9 - 24}}{12}$

$x = \frac{- 3 \pm \sqrt{- 15}}{12}$

$x = \frac{- 3 \pm i \sqrt{15}}{12}$

Roots:

$x = \frac{- 3 + i \sqrt{15}}{12} ,$$\frac{- 3 - i \sqrt{15}}{12}$