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

##### 1 Answer
Dec 31, 2017

Required Solutions are:

$\textcolor{b l u e}{x = \frac{1}{2} \mathmr{and} x = 0.5}$

$\textcolor{b l u e}{x = - 1}$

#### Explanation:

We are given

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

We need to use the quadratic formula to find the roots, real or imaginary.

The algebraic expression on the Right-hand-Side(RHS) can be simplified to ...

$\textcolor{b l u e}{- \left({x}^{2} - 2 x + 1\right) - {x}^{2} - 3 x + 2}$

$\textcolor{b l u e}{\Rightarrow - {x}^{2} + 2 x - 1 - {x}^{2} - 3 x + 2}$

$\textcolor{b l u e}{\Rightarrow - 2 {x}^{2} - x + 1}$

Now we have

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

The General Form of a Quadratic Equation is

$\textcolor{red}{a} {x}^{2} + \textcolor{red}{b} x + \textcolor{red}{c} = 0$

The Quadratic Formula for finding the roots of a quadratic equation is given by

$\textcolor{b l u e}{x = \frac{\left(- b\right) \pm \sqrt{{b}^{2} - 4 a c}}{2 a} , a \ne 0}$

$\textcolor{g r e e n}{{b}^{2} - 4 a c}$ is referred to as the Discriminant

When $\textcolor{g r e e n}{\left({b}^{2} - 4 a c\right) = 0}$ there exists One Real Root

When $\textcolor{g r e e n}{\left({b}^{2} - 4 a c\right) > 0}$ there exists Two Real Roots

When $\textcolor{g r e e n}{\left({b}^{2} - 4 a c\right) < 0}$ there exists Two Complex Roots

For our Quadratic Equation

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

We observe that

$\textcolor{red}{a} = - 2$

$\textcolor{red}{b} = - 1$

$\textcolor{red}{c} = 1$

Substitute these values in the Quadratic Formula

color(blue)(x=((-b)+- sqrt(b^2 - 4ac))/(2a)

We get

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

On simplification we get,

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

$x = \frac{1 \pm \sqrt{9}}{- 4}$

We observe that, the discriminant $\left({b}^{2} - 4 a c\right) > 0$

Hence, we will have Two Real Roots

$x = \frac{1 \pm 3}{-} 4$

rArr x= (1+3)/-4; x = (1-3)/-4

rArr x= 4/-4; x = -2/-4

rArr x= -1; x = 1/2

Hence our two real roots are

$\textcolor{b l u e}{x = \frac{1}{2} \mathmr{and} x = 0.5}$

$\textcolor{b l u e}{x = - 1}$

Hope this helps.