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

1 Answer
Tamir E.
Dec 27, 2017

x_1=3+2sqrt2ix1=3+2√2i
x_2=3-2sqrt2ix2=3−2√2i

Explanation:

y=-(x-4)^2-2x-1=y=−(x−4)2−2x−1=
=-(x^2-8x+16)-2x-1==−(x2−8x+16)−2x−1=
=-x^2+8x-16-2x-1==−x2+8x−16−2x−1=
=-x^2+6x-17=−x2+6x−17

Let y=0y=0
=>⇒
-x^2+6x-17=0 iff a=-1, b=6, c=-17−x2+6x−17=0⇔a=−1,b=6,c=−17
=>⇒
x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}=x1,2=−b±√b2−4ac2a=
={-6+-sqrt{6^2-4*(-1)*(-17)}}/{2(-1)}==−6±√62−4⋅(−1)⋅(−17)2(−1)=
={-6+-sqrt{36-68}}/{-2}==−6±√36−68−2=
={-6+-sqrt{-32}}/{-2}==−6±√−32−2=
={-6+-sqrt{-(16*2)}}/{-2}==−6±√−(16⋅2)−2=
={-6+-4sqrt2i}/{-2}==−6±4√2i−2=
={cancel(-2)(3+-2sqrt2i)}/{cancel(-2)}=
=3+-2sqrt2i
=>
x_1=3+2sqrt2i
x_2=3-2sqrt2i

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