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

Feb 10, 2018

$y = \frac{5 + \sqrt{79}}{6}$ and $\frac{5 - \sqrt{79}}{6}$

#### Explanation:

First things first, we need to get this into standard form, $a {x}^{2} + b x + c$

expand

$\left(2 x - 1\right) \left(2 x - 1\right) + 2 \left(x + 1\right) \left(x - 4\right)$

distribute

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

$4 {x}^{2} - 4 x - 1 + 2 {x}^{2} - 6 x - 8$

combine like-terms

$\textcolor{\mathmr{and} a n \ge}{6} {x}^{2} + \textcolor{b l u e}{- 10} x + \textcolor{\pi n k}{- 9}$

The quadratic formula is $\frac{- b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 \times a \times c}}{2 a}$

with $a = \textcolor{\mathmr{and} a n \ge}{6}$, $b = \textcolor{b l u e}{- 10}$, and $c = \textcolor{\pi n k}{- 9}$

$\frac{- \textcolor{b l u e}{- 10}}{2 \times \textcolor{\mathmr{and} a n \ge}{6}} \pm \frac{\sqrt{{\left(\textcolor{b l u e}{- 10}\right)}^{2} - 4 \times \textcolor{\mathmr{and} a n \ge}{6} \times \textcolor{\pi n k}{- 9}}}{2 \times \textcolor{\mathmr{and} a n \ge}{6}}$

simplify

$\frac{10}{12} \pm \frac{\sqrt{100 + 216}}{12}$

$\frac{5}{6} \pm \frac{2 \sqrt{79}}{12}$

$\frac{5}{6} \pm \frac{\sqrt{79}}{6}$

$y = \frac{5 + \sqrt{79}}{6}$ and $\frac{5 - \sqrt{79}}{6}$

Feb 10, 2018

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

color(blue)(y=6x^2-10x-7

Roots : color(blue)(x_1 = [5-sqrt(67)]/6, " " x_2 = [5+sqrt(67)]/6 " " (or)

${x}_{1} = - 0.53089 , \mathmr{and} {x}_{2} = 2.19756$

#### Explanation:

Given:

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

color(green)(Step.1

We will expand and simplify

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

as shown below:

First, consider color(blue)((2x-1)^2

Using the algebraic identify,

color(brown)((a-b)^2 -= a^2 - 2ab - b^2,

we can expand as shown:

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

$\Rightarrow 4 {x}^{2} - 4 x + 1 \text{ }$ Expression.1

color(green)(Step.2

Next, consider

color(blue)(2 (x + 1) (x - 4)

We can factor them out to obtain

$\Rightarrow 2 \cdot \left({x}^{2} - 4 x + x - 4\right)$

$\Rightarrow 2 \cdot \left({x}^{2} - 3 x - 4\right)$

$\Rightarrow 2 {x}^{2} - 6 x - 8 \text{ }$ Expression.2

color(green)(Step.3

We will add our Expression.1 and Expression.2 to get $y$

$\left(4 {x}^{2} - 4 x + 1\right) + \left(2 {x}^{2} - 6 x - 8\right)$

$\Rightarrow 4 {x}^{2} - 4 x + 1 + 2 {x}^{2} - 6 x - 8$

$\Rightarrow 6 {x}^{2} - 10 x - 7$

Now, we are in a position to re-write our Polynomial as

color(blue)(y=f(x)=6x^2-10x-7

color(blue)(6x^2-10x-7=0 is our Quadratic Equation,

and we proceed to find the roots.

color(green)(Step.4

Now that we have a Quadratic Equation in the Standard Form

$a {x}^{2} + b x + c = 0$, where

a = 6; b = (-10) and c = (-7)

We now use the Quadratic Formula to find the roots

color(red)(Root_1,_2 = -b+- sqrt(b^2-4*a*c)/(2.a)

rArr [-(-10)+-sqrt((-10)^2-4(6)(-7))]/(2(6)

$\Rightarrow \frac{10 \pm \sqrt{100 - 4 \left(- 42\right)}}{12}$

$\Rightarrow \frac{10 \pm \sqrt{100 + 168}}{12}$

$\Rightarrow \frac{10 \pm \sqrt{268}}{12}$

$\Rightarrow \frac{10}{12} \pm \frac{\sqrt{268}}{12}$

$\Rightarrow \frac{10}{12} \pm \frac{\sqrt{4 \cdot 67}}{12}$

$\Rightarrow \frac{10}{12} \pm \frac{\sqrt{4}}{12} \cdot \frac{\sqrt{67}}{12}$

$\Rightarrow \frac{5}{6} \pm \frac{2}{12} \cdot \frac{\sqrt{67}}{12}$

$\Rightarrow \frac{5}{6} \pm \frac{2 \sqrt{67}}{12}$

rArr 5/6+-(cancel 2sqrt(67))/(cancel 12( color(red)(6))

$\Rightarrow \frac{5}{6} \pm \frac{\sqrt{67}}{6}$

$\Rightarrow \frac{5 \pm \sqrt{67}}{6}$

Hence, we can write our roots as

color(blue)(Root_1 = [5-sqrt(67)]/6, " " Root_2 = [5+sqrt(67)]/6 " " (or)

$R \infty {t}_{1} = - 0.53089 , \mathmr{and} R \infty {t}_{2} = 2.19756$

Please observe that these are our required real roots, as the discriminant color(blue)((b^2-4*a*c)>0

color(green)(Step.5

We will verify our solutions using GeoGebra graphing software 