# How do you find the roots, real and imaginary, of y= x^2 - 6x +3 -(2x- 3 )^2  using the quadratic formula?

## $y = {x}^{2} - 6 x + 3 - {\left(2 x - 3\right)}^{2} = 0$

Jul 27, 2018

$\implies D = - 36 < 0 \implies$the roots are imaginary numbers.

#### Explanation:

Here,

$y = {x}^{2} - 6 x + 3 - {\left(2 x - 3\right)}^{2}$

$y = {x}^{2} - 6 x + 3 - \left(4 {x}^{2} - 12 x + 9\right)$

$\implies y = {x}^{2} - 6 x + 3 - 4 {x}^{2} + 12 x - 9$

$\implies y = - 3 {x}^{2} + 6 x - 6$

If $y = 0$ ,then $- 3 {x}^{2} + 6 x - 6 = 0 \text{ is a quadratic equation .}$

Comparing with $a {x}^{2} + b x + c = 0$ ,we get

$a = - 3 , b = 6 \mathmr{and} c = - 6$

So, $\text{the "color(blue)"Discriminant}$ of the quadratic eqn. is :

$\textcolor{b l u e}{D = {b}^{2} - 4 a c} = {6}^{2} - 4 \left(- 3\right) \left(- 6\right) = 36 - 72$

$\implies D = - 36 < 0 \implies$the roots are imaginary numbers.

$i . e . x \in \mathbb{C} \to$ complex roots.