How do you find the roots, real and imaginary, of y=-3x^2 -6x+4(x -1)^2  using the quadratic formula?

Oct 31, 2017

color(magenta)(x=7+3sqrt5

orcolor(magenta)(x=7-3sqrt5

Explanation:

If the question excludes the"y=" then:-

$\therefore - 3 {x}^{2} - 6 x + 4 {\left(x - 1\right)}^{2}$

$\therefore = - 3 {x}^{2} - 6 x + 4 \left({x}^{2} - 2 x + 1\right)$

$\therefore = - 3 {x}^{2} - 6 x + 4 {x}^{2} - 8 x + 4$

$\therefore = {x}^{2} - 14 x + 4$

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

$\therefore x = \frac{- \left(- 14\right) \pm \sqrt{{\left(- 14\right)}^{2} - 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$

$\therefore x = \frac{14 \pm \sqrt{196 - 16}}{2}$

$\therefore x = \frac{14 \pm \sqrt{180}}{2}$

$\therefore x = \frac{14 \pm \sqrt{2 \cdot 2 \cdot 3 \cdot 3 \cdot 5}}{2}$

$\therefore x = \frac{14 \pm 6 \sqrt{5}}{2}$

$\therefore x = {\cancel{14}}^{7} / {\cancel{2}}^{1} \pm {\cancel{6}}^{3} \frac{\sqrt{5}}{\cancel{2}} ^ 1$

:.color(magenta)(x=7+3sqrt5 or color(magenta)(x=7-3sqrt5