How do you find the roots, real and imaginary, of y=-(x -4 )^2-2x-1y=(x4)22x1 using the quadratic formula?

1 Answer
Dec 27, 2017

x_1=3+2sqrt2ix1=3+22i
x_2=3-2sqrt2ix2=322i

Explanation:

y=-(x-4)^2-2x-1=y=(x4)22x1=
=-(x^2-8x+16)-2x-1==(x28x+16)2x1=
=-x^2+8x-16-2x-1==x2+8x162x1=
=-x^2+6x-17=x2+6x17

Let y=0y=0
=>
-x^2+6x-17=0 iff a=-1, b=6, c=-17x2+6x17=0a=1,b=6,c=17
=>
x_{1,2}={-b+-sqrt{b^2-4ac}}/{2a}=x1,2=b±b24ac2a=
={-6+-sqrt{6^2-4*(-1)*(-17)}}/{2(-1)}==6±624(1)(17)2(1)=
={-6+-sqrt{36-68}}/{-2}==6±36682=
={-6+-sqrt{-32}}/{-2}==6±322=
={-6+-sqrt{-(16*2)}}/{-2}==6±(162)2=
={-6+-4sqrt2i}/{-2}==6±42i2=
={cancel(-2)(3+-2sqrt2i)}/{cancel(-2)}=
=3+-2sqrt2i
=>
x_1=3+2sqrt2i
x_2=3-2sqrt2i