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

1 Answer
Aug 5, 2016

Answer:

The roots are #x= 1/2# and #x=1/3#

Explanation:

We begin with #y=2x^2-x+(2x-1)^2#. Our first step is to simplify our equation until it cannot be reduced any further. Then we move on to the quadratic formula.

#y=2x^2-x+(2x-1)^2# can be changed to #y=2x^2-x+((2x-1)*(2x+1))#. That becomes #y=2x^2-x+4x^2-2x-2x+1# or #y=2x^2-x+4x^2-4x+1#. Now we combine like terms like #2x^2+4x^2# and #-x-5x#. That gives us #y=6x^2-5x+1#.

Now we can move on to the quadratic equation, which is #(-b+-sqrt(b^2-4*a*c))/(2a)# where those variables come from #ax^2+bx+c#. In our case #a=6#, #b=-5#, and #c=1#. Now we just plug in our value to the formulla: #(5+-sqrt(-5^2-4*6*1))/(2*6)#. This can be simplified to #(5+-sqrt(25-24))/(12)# or #(5+-1)/12#. That becomes #4/12# or #6/12#, which can be sipmlified to #1/3# and #1/2#, respectivley. Those are our roots, and we can graph the original equation and look at the #x#-intercepts (roots).

graph{y=6x^2-5x+1}

And they are! Nice work!