How do you find the roots, real and imaginary, of #y=-5x^2+7x+2(2x-1)^2 # using the quadratic formula?

1 Answer
May 15, 2016

Answer:

The roots are #(1+-isqrt23)/6#

Explanation:

We begin with #y=-5x^2+7x+2(2x-1)^2#. In order to use the quadratic formula, we need to get this equation into standard form, which means that the format needs to look like this: #ax^2+bx+c#.

So, in order to change the form of our current equation, we first need to expand everything that we can. Let's start with #(2x-1)^2#. This becomes #(2x-1)(2x-1)#, which is #4x^2-4x+1#. Now we have #-5x^2+7x+2(4x^2-4x+1)#/ By multiplying out the #2#, we arrive at #-5x^2+7x+8x^2-8x+2#.

From here, we just need to combine like terms. #-5x^2+8x^2# gives us #3x^2#, while #7x-8x# equals #-x#. The #2# is left alone.

So now we have the following: #y=3x^2-x+2#, which is in standard form, meaning that we are good to go for using the quadratic formula!

#(-b+-sqrt(b^2-4*a*c))/(2*a)#

#color(blue)(a=3)#
#color(green)(b=-1)#
#color(red)(c=2)#

We just need to substitute the #a#s and #b#s and #c#s for their numerical value.

The formula now becomes

#(-(color(green)(-1))+-sqrt((color(green)(-1))^2-4*(color(blue)(3))*(color(red)(2))))/(2(color(blue)(3)))#

Simplifying that we have #(1+-sqrt(1-24))/6# or #(1+-isqrt(23))/6#.

That is the imaginary root for #y=-5x^2+7x+2(2x-1)^2#. Nice job.