Question

How do you find the roots, real and imaginary, of `y=-(x -4 )^2-2x-1` using the quadratic formula?

Answer 1

`x_1=3+2sqrt2i`
`x_2=3-2sqrt2i`

Explanation:

`y=-(x-4)^2-2x-1=`
`=-(x^2-8x+16)-2x-1=`
`=-x^2+8x-16-2x-1=`
`=-x^2+6x-17`

Let `y=0`
`=>`
`-x^2+6x-17=0 iff a=-1, b=6, c=-17`
`=>`
`x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}=`
`={-6+-sqrt{6^2-4*(-1)*(-17)}}/{2(-1)}=`
`={-6+-sqrt{36-68}}/{-2}=`
`={-6+-sqrt{-32}}/{-2}=`
`={-6+-sqrt{-(16*2)}}/{-2}=`
`={-6+-4sqrt2i}/{-2}=`
`={cancel(-2)(3+-2sqrt2i)}/{cancel(-2)}=`
`=3+-2sqrt2i`
`=>`
`x_1=3+2sqrt2i`
`x_2=3-2sqrt2i`