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

Apr 12, 2017

There are no imaginary roots as the graph has x-intercepts.

$\text{ } x \approx - 4.45 \mathmr{and} x \approx 0.45$ to 2 decimal places

#### Explanation:

First we need to change the given equation so that it is in the form $y = a {x}^{2} + b x + c$

Square the brackets

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

$y = 3 {x}^{2} - 2 {x}^{2} + 4 x - 2$

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

To determine the roots set $y = 0$

$0 = {x}^{2} + 4 x - 2$

So we have: $a = 1 \text{; "b=4"; } c = - 2$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ } \to x = \frac{- 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$

$\text{ } x = - 2 \pm \frac{\sqrt{16 + 8}}{2}$

$\text{ } x = - 2 \pm \frac{\sqrt{24}}{2}$

But $24 = 2 \times 3 \times 4$ so $\sqrt{24} = 2 \sqrt{6}$

$\text{ } x = - 2 \pm \frac{2 \sqrt{6}}{2}$

$\text{ } x = - 2 \pm \sqrt{6}$

$\text{ } x \approx - 4.45 \mathmr{and} x \approx 0.45$