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

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Answer 1 · 2017-12-11T14:01:27

#x=4-sqrt(2)#, #4+sqrt(2)#
Check below for my work!

You need to put it this in a form that's usable with the quadratic formula. Generally you want an equation in the form of #ax^2+bx+c#.

Let's work with the equation a bit. To start, we have two binomials to the power of two. We can expand these with multiplication.

#-2color(blue)((x-3)^2)+color(red)((-x-2)^2)#

Binomnial 1
#color(blue)((x-3)^2)#
#color(blue)((x-3)(x-3))# Multiply these together using the FOIL method.
#color(blue)(x^2-6x+9)#

Binomial 2
#color(red)((-x-2)^2)#
#color(red)((-x-2)(-x-2))# Again, use the FOIL method.
#color(red)(x^2-4x+4)#

Great! Now we're left with #-2(x^2-6x+9)+(x^2-4x+4)#.

Now we can multiply the first polynomial by #-2#.

#-2(x^2-6x+9)#
#-2x^2+12x-18#

That leaves us with #(-2x^2+12x-18)+(x^2-4x+4)#

Now we simply combine like terms.

#(-2x^2+x^2)+(12x-4x)+(-18+4)#
#-x^2+8x-14#

Pulling from our simplified polynomial, we can plug these values into the quadratic formula.

#a=-1#
#b=8#
#c=-14#

Finally, we solve!

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

#x=(-(8)+-sqrt((8)^2-4(-1)(-14)))/(2(-1))#

#x=(-8+-sqrt(8))/-2#

#x=(-8+-2sqrt(2))/-2#

*I'm splitting the #+-# up here. The #+# equation is in #color(blue)("blue")# while the #-# equation is in #color(red)("red")#.

#color(blue)(x=(-8+2sqrt(2))/-2)#

#color(blue)(x=4-sqrt(2))#

#color(red)(x=(-8-2sqrt(2))/-2)#

#color(red)(x=4+sqrt(2))#

Since these can't be simplified any further, these are the real roots. There are no imaginary roots.