How do you find the roots, real and imaginary, of #y=-6x^2 + 24x +24(x/2-1)^2 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y=-6x^2 + 24x +24(x/2-1)^2        # using the quadratic formula?

Julia Adams · Accepted Answer

No roots

First,we need to get this equation into Standard Form #y = -6x^2+24x+24(x/2-1)^2# #y = -6x^2+24x+24(x^2/4-x+1)# #y = cancel(-6x^2)+cancel(24x)+cancel(6x^2)-cancel(24x)+24# #y=24# Well, I guess we don't need to use any quadratic formula for this There are no roots for this, see graph{y=24x/x} It never crosses the #x-#axis

Genesis Allen · Answer

undefined To find the roots of the quadratic equation $ y = -6x^2 + 24x + 24\left(\frac{x}{2} - 1\right)^2 $ using the quadratic formula, first, express the equation in standard quadratic form, $ ax^2 + bx + c = 0 $.

$ y = -6x^2 + 24x + 24\left(\frac{x}{2} - 1\right)^2 $
Expand the squared term:
$ y = -6x^2 + 24x + 24\left(\frac{x^2}{4} - x + 1\right) $
Distribute 24 into the parentheses:
$ y = -6x^2 + 24x + 6x^2 - 24x + 24 $
Combine like terms:
$ y = 24 $

Now, apply the quadratic formula:
$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

In this equation, $ a = -6 $, $ b = 24 $, and $ c = 24 $.

Plug these values into the quadratic formula and solve for $ x $:
$ x = \frac{{-24 \pm \sqrt{{24^2 - 4(-6)(24)}}}}{{2(-6)}} $
$ x = \frac{{-24 \pm \sqrt{{576 + 576}}}}{{-12}} $
$ x = \frac{{-24 \pm \sqrt{{1152}}}}{{-12}} $
$ x = \frac{{-24 \pm 8\sqrt{18}}}{{-12}} $
$ x = 2 \pm \frac{{2\sqrt{18}}}{{3}} $

So, the roots of the equation are:
$ x = 2 + \frac{{2\sqrt{18}}}{{3}} $ and $ x = 2 - \frac{{2\sqrt{18}}}{{3}} $