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

1 Answer
Dec 11, 2017

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

Explanation:

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

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.