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

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Answer 1 · 2017-08-25T06:24:24

#x = 7/2+-sqrt(85)/2#

By roots, I will assume you mean zeros.

First expand out the quadratic, then combine terms to bring it into standard form:

#2x^2-x-(x+3)^2 = 2x^2-x-(x^2+6x+9)#

#color(white)(2x^2-x-(x+3)^2) = 2x^2-x-x^2-6x-9#

#color(white)(2x^2-x-(x+3)^2) = x^2-7x-9#

This is in the form:

#ax^2+bx+c#

with #a=1#, #b=-7# and #c=-9#

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-7))^2-4(color(blue)(1))(color(blue)(-9)) = 49+36 = 85#

Since #Delta > 0#, this quadratic has two distinct real zeros, but since #85# is not a perfect square, those zeros are irrational.

We can use the quadratic formula to find them:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (7+-sqrt(85))/2#

#color(white)(x) = 7/2+-sqrt(85)/2#